j^cto |9orfe fetatc College of agriculture at Cornell ®nibers(itp Stbata, J9. 1?. Hiftrarp Cornell University Library QA 273.W6 Choice and chance; an elementary treatise 3 1924 002 940 280 M ^ Cornell University VM Library The original of tliis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924002940280 CHOICE AND CHANCE, * AN ELBMENTAET TEEATISE ON PERMUTATIONS, COMBINATIONS, AND PROBABILITY, ■WITH 640 EXEKCISBS; WILLIAM ALLEN WHITWORTH, M.A. LATE EELLOW OF ST JOHS's OOHLEGE, OAMBEIDOE. FOURTH EDITION, ENLARGED. CAMBRIDGE : DEIGHTON, BELL AND CO. LONDON: GEOEGE BELL AND SONS. 1886 [All rights reserved.] ffiambtitfge: EEINTED BY 0. J. CLAY, M.A. AND SONS, AT THE UNIVBESITY PBESS. PREFACE TO THE FOURTH EDITION. The " two Chapters of Arithmetic " of the original edition appear as Chapters I. and VI. of the present volume. These two Chapters (together with a considerable part of Chapter IX.) will be intelligible to a reader who has no knowledge of mathematics, and will put him into possession of the principles upon which simple questions of probability are determined. The other Chapters assume a knowledge of Algebra. Questions requiring the application of the Integral Calculus are not included in the book, which only fulfils its title to be an Elementary Treatise. Important additions have been made since the publication of the third edition. I refer especially to the discussion of conditions of precedence in Chapter V. (the substance of which I published in the Messenger of Mathematics in 1878), and the interesting propositions L. and Li. which connect the average frequency of an event, with the probability of its happening in a given period. IV PEEFACE, I have given at pages 168 and 194 somewhat fuller hints as to the application of the theory of Chance to questions of expectation of life ; not attempting to furnish the Actuary with the apparatus of his craft, but endeavouring rather to exhibit and illustrate the principles which must justify his rules. I have increased the number . of exercises at the end of the Volume from 300 to 640, and I am preparing solutions of the whole series for immediate publication. St John's Vioabaqe, Hammersmith, 24 July, 1886. CONTENTS. CHOICE. CHAPTERS I. AND II. PAGE 59 60 61 62 Multiplication of Choice. Eule i. p. 8, or Prop. i. . . . Oontirmed-nmltiplication of Choice. Eule ii. p. 13, or Prop. n. . Permutatione, all the things together. Eule m. p. 18, or Prop. m. Permutations, a given number together. Eule rv. p. 23, or Prop. it. Division into two classes. Eule v. p. 25, or Prop, v 63 Division into a number of classes. Eule vi. p. 27, or Prop. yi. . .64 Permutations, some of the things being alike. Eule vn. p. 33, or Prop. VII. . . . . . . . . . . .65 Combinations. Eules viii. and ix. pp. 35, 36, or Prop. viii. ... 66 Combinations ■with repetition. Eule r. p. 46, orProp. ix. . . .68 Permutations with repetition. Eule'ii. p. 52, or Prop. i. . . .71 Total number of combinatiOJis. Eule xn. p. 53, or Prop. xi. . . 71 The same, some of the things being alike. Eule xiii. p. 57, or Prop. xii. 72 The Binomial Theorem. Prop. xin. ,72 Occurrences of stated conditions among, the various ways of performing an operation. Prop, xiv 73 EXCUESUS. On certain Series ... 77 VI CONTENTS. CHAPTEK III. PAGE Distribution of a given number of different things into different groups. Props. XV. and xvi 89 The same, into indifferent groups. Prop, xvii 90 Distribution out of a given number of different things into different groups. Props, xviii. and xix 90, 92 The same, into indifferent groups. Prop, xx 93 Distribution of different things into different parcels. Props, xxi. and XXII 93 Distribution of diflerent things into indifferent parcels. Props, xxiii. and XXIV 94, 95 Distribution of indifferent things into different parcels. Prop. xxv. and XXVI 96, 97 The same, when the number in each parcel is limited. Props, xxvii. and XXVIII 98 Distribution of indifferent things into indifferent parcels. Props, xxix. and XXX. 100, 104 Table of the number of j'-partitions of re 106 CHAPTER IV. Sub-factorial notation. 107 Number of derangements of a series of elements, so that no element remains in situ. Prop, xxxi 109 The same, so that no original sequence of two terms remains. Props. xxxii. and xxxiii 110 m Derangements of terms in circular order. Prop, xxxiv. . . . 112 Table of factorials and sub-factorials 114 CHAPTER V. Questions involving the consideration of priority. . . . .115 Order of gains and losses, the losses never exceeding the gains. Props. XXXV. to xxxviii 117 122 The case when the excess is to be limited to a certain amount. Prop. xxxix. 123 CONTENTS. Vll CHANCE. CHAPTERS VI. AND VII. PAGE Complementary probabilities. Eule i. p. 132, or Prop. xl. . . . 188 Eelatiou between probability and odds. Eule ii. p. 134, or Prop. xli. 188 Fundamental Proposition in the calculation of chances. Eules iii. and IV. pp. 135, 137, or Props, xlii. and xliii 189 Expectation. Eule v. p. 148, or Prop. xLiv. 189 Multiplication of chances of independent events. Eule vi. p. 151, or Prop. xLv 190 Multiplication of chances of events not independent. Eule vn. p. 153, or Prop. XLVi. 192 Continued-multiplication of chances. Eule vni. p. 159, or Prop, xlvii. 192 Vital Statistics p. 168 Inverse probabilities. Eule ix. p. 171, or Prop, xlviii. . . ■ 192 Credibility of testimony, p. 179 Game of Whist, pp. 182—187 Notes on Population and Expectation of life 194 CHAPTER VIII. Successive trials Average number of trials before succeeding. Prop. xlix. Eelation of chance to average. Props. ±<. and li. Chance of consecutive successes. Props, lii. and liii. Average number of trials before consecutive successes. Prop, liv, . 198 . 200 201, 202 . 203 . 205 CHAPTER IX. The disadvantage of gambling Insurance the reverse of gambling 207 213 VIU CONTENTS. FAQE Effect of the repetition of a fair wager. Prop. lv. .... 216 Effect of the repetition of a wager at odds. Prop. LVi 218 Repetition of a fair wager on a scale proportioned to the speculator's means. Prop. lvii. 222 Price which a man of limited means may pay for a given chance of a prize. Prop. Lvni. 224 The general case of a lottery with prizes of different value. Props. Lix. andLx 226, 229 The Petersburg problem. Prop, lxi 231 CHAPTER X. Oeometrical Eepresentation of Chances 237 EXERCISES. On Choice 244 On Chance 262 Answers to the Exercises 295 ERRATA. Page 81 lines 11, 12 the afSx n-p-1 ought to be j» - ra+ 1. Page 81 line 5 from bottom for Glzl read 0"+*-'. Page 108 lines 3, 4 from bottom , odd, , even, for read even, odd. Page 111 footnote for greatest integer in read integer nearest to. CHOICE. CHAPTER I. THE THEORY OF PERMUTATIONS AND COMBINATIONS TREATED ARITHMETICALLY. We have continually to make our choice among different courses of action open to us, and upon the discretion with which we make it, much may depend. Of this discretion a higher philosophy must treat, and it is not to be supposed that Arithmetic has anything to do with it ; but it is the province of Arithmetic, under given circumstances, to measure the choice which we have to exercise, or to determine precisely the number of courses open to us. Suppose, for instance, that a member is to be returned to parliament for a certain borough, and that four candidates present themselves. Arithmetic has nothing to do with the manner in which we shall exercise our privilege as a voter, which depends on our discretion in judging the qualifications of the different candidates ; but it belongs to Arithmetic, as the science of counting and calculation, to tell us that the number of ways in which Cif we vote at all) we can exercise our choice, is four. w. 1 2 CHOICE. The operation is, indeed, in this case so simple that we scarcely recognise its arithmetical character at all; but if we pass on to a more complicated case, we shall observe that some thought or calculation is required to determine the number of courses open to us: and thought about numbers is Arithmetic. Suppose, then, that the borough has to return two mem- bers instead of one. And still suppose that we have the same four candidates, whom we will distinguish by names, as A, B, C, B. If we try to note down all the ways in which it is possible for us to vote, we shall find them to be six in number; thus we may vote for any of the following : — A and B, A and C, B and C, A and D, B and D, C and D. But we can hardly make this experiment without perceiving that the resulting number, six, must in some way depend arithmetically upon the number of candidates and the number of members to be returned, or without suspecting that on some of the principles of arithmetic we ought to be able to arrive at that result without the labour of noting all the possible courses open to us, and then counting them up; a labour which we may observe would be very great if ten or twelve candidates offered themselves, instead of four. In the present chapter we shall establish and explain the principles upon which such calculations are made arith- metically. It will be found that they are very simple in nature as well as few in number. In the next four chap- ters we shall treat the subject somewhat more largely by algebraical methods; but the reader who is unacquainted with algebra may pass over those chapters, and proceed with the DIFFERENT SELECTIONS. 3 first chapter on Chance (Chap, VI.), in which he will find the principles of Choice applied arithmetically to the solutions of problems in Probability, a subject of great interest and some practical importance. We found, by experiment or trial, that there were six ways of voting for two out of four candidates. So we may say that, out of any four given articles, six selections of two articles may be made. But we call special attention to the sense in which we use the words "six selections." We do not mean that a man can select two articles, and having taken them can select two more, and then two more, and so on till he has made six selections altogether ; for it is obvious , that the four articles would be exhausted by the second selection; but when we speak of six selections being possible, we mean that there are six different ways of making one selection, just as among four candidates there are six ways of selecting two to vote for. This language may appear at first to be arbitrary and unnecessary, but as we proceed with the subject we shall find that it simplifies the expression of many of our results. In making the selection of two candidates out of four, in the case just considered, it was immaterial which of the two selected ones we took first ; the selection of A first, and then B, was to every intent and purpose the same thing as the selection of B first, and then A. But if we alter the question a little, and ask in how many ways a society can select a president and vice-president out of four candidates for office, the order of selection be- comes of importance. To elect A and B as president and vice-president respectively, is not the same thing as to elect B and A for those two offices respectively. Hence there are twice as many ways as before of making the election, viz. — 1—2 4 CHOICE, A and B, A and G, B and C, A and D, B and B, G and -D, B and ,4, (7 and A, G and 5, D and A, D and 5, D and (7. So if four articles of any kind are given us, there will be twelve ways of choosing two of them in a particular order ; or, as we may more briefly express it, out of four given articles, twelve arrangements of two articles can be made. But it must be observed that the same remarks apply here, which we made on the use of the phrase " six selections " lon p. 3. We do not mean that twelve arrangements or six selections can be successively made; but that if one arrangement or one selection of two articles have to be made out of the four given articles, we have the choice of twelve ways of making the arrangement, and of six ways of making the selection. We may give the following formal definitions of the words selection and arrangement, in the sense in which we have used them : — Def. I. — A selection (or combination) of any number of articles, means a parcel of that number of articles classed together, but not regarded as having any particular order among themselves. Def. II. — An arrangement (or permutation) of any number of articles, means a group of that number of articles, not only classed together, but regarded as having a parti- cular order among themselves. Thus the six groups, — ABG, BGA, GAB, AGB, BAG, GBA, are all the same selection (or combination) of three letters. SELECTIONS AND AEEANGEMENTS. 6 but they are all different arrangements (or permutations) of three letters. So, out of the four letters A, B, C, D, we can make four selections of three letters, viz. — BCD, CD A, BAB, ABC; but out of the same four letters we can make twenty-four arrangements of three letters, viz. — BCD, BDG, GDB, GBD, DBG, DGB, GDA, GAD, DAG, DGA, AGD, ADG, DAB, DBA, ABD, ADB, BDA, BAD, ABG, AGB, BGA, BAG, GAB, GBA. It should be noticed that the results which we have just stated are often enunciated in slightly different phraseology. There is no difference of meaning whether we say (1) The number of selections (or combinations) of 3 things out of 4 ; or (2) The number of ways of selecting 3 things out of 4; or (3) The number of combinations of 4 things taken 3 at a time (or taken 3 together). Similarly we may speak indifferently of (1) The number of arrangements of 3 things out of 4; or (2) The number of orders of arranging 3 things out of 4; or (3) The number of permutations (or arrangements) of 4 things taken 3 at a time (or taken 3 together). CHOICE. Having thus explained the language we shall have to employ, we may now proceed to establish the principles on which all calculations of choice must be founded. The great principle upon which we shall base all our reasoning may be stated as follows : — If one thing cam, he done in a given number of different ways, and then another thing in another given number of different ways, the number of different ways in which both things can be done is obtained by multiplying together the two given numbers. "We shall first illustrate this principle, and then proceed to prove it. Suppose we have a box containing five capital letters, A, B, G, D, E, and three small letters, x, y, z. A B ' G B E X y z The number of ways in which we can select a capital letter out of the box is jive ; the number of ways in which we can select a small letter is three ; therefore, by the principle we have just stated, the number of different ways in which we can select a capital letter and a small one is fifteen, which we find on trial to be correct, all the possible selections being as follows : — Ax, Bx, Gx, Bx, Ex, Ay, By, Oy, Dy, Ey, As, Bz, Gz, Bz, Ez. Again, suppose there are four paths to the top of a SELECTIONS AND ARRANGEMENTS. 7 mountain, the principle asserts that we have the choice of sixteen ways of ascending and descending. For there are 4 ways up, 4 ways down, and 4 X 4 = 16. We can verify this : for if P, Q, E, 8 be the names of the four paths, we can make our choice among the following sixteen plans, the first-mentioned path being the way up, and the second the way down : — P and P, P and Q, P and R, ■ P and S, Q and P, Q and Q, Q and R, Q and S, R and P, R and Q, R and R, B and 8, S and P, S and Q, S and R, 8 and 8. Or, if we had desired to ascertain what choice we had of going up and down by different paths, we might still have applied the principle, reasoning thus : There are four ways of going up, and when we are at the top we have the choice of three ways of descending (since we are not to come down by the same path that brought us up). Hence the number of ways of ascending and descending is 4 X 3, or twelve. These twelve ways will be obtained from the sixteen de- scribed in the former case, by omitting the four ineligible ways, P and P, Q and Q, R and R, 8 and 8. The foregoing examples will suffice to illustrate the meaning and application of our fundamental proposition. We will now give a formal proof of it. We shall henceforth refer to it as Rule I. CHOICE. RULE I. If one thing can he dam in a given number of different ways, and (when it is done in any way) another thing can be done in another given number of different ways, then the number of different ways in which the two things can he done is the product of the two given numbers. For let A, B, G, D, E, &c. represent the different ways in which the first thing can be done (taking as many letters as may be necessary to represent all the different ways), and similarly let a, h, c, d, &c. represent the different ways of doing the second thing. Then, if we form a table as below, having the letters A, B, G, D, E, &c. at the head of the several columns, and the letters a, h, c, d, &c. at the end of the several horizontal rows, we may regard each square in the table as representing the case, in which the first thing is WATS OF DOING THE FIB3T THINS. A B G D E F G &c. ■a 5 w 1 i a b c « 1 t 1 d &c. ' CHOICE IN SUCCESSIVE OPERATIONS. 9 done in the way marked at the head of the column in which the square is taken, and the second thing in the way marked at the end of the row. Thus the square marked with the asterisk (*) will denote the case in which the first thing is done in the way which we called 0, and the second thing in the way which we called b I and the square marked with the dagger ("I-) will denote the case in which the first thing is done in the way E, and the second in the way c ; and so on. Now it will be readily seen that all the squares represent different cases, and that every case is represented by some square or other. Hence the number of possible cases is the same as the number of squares. But there are as many columns as there are ways of doing the first thing, and each column contains as many squares as there are ways of doing the second thing. Therefore the number of squares is the product of the number of ways of doing the two several things, and therefore this product expresses also the whole number of possible cases, or the whole number of ways in which the two events can be done. This proves the rule. Question 1. A cabinet-maker has twelve patterns of chairs and five patterns of tables. In how many ways can he make a chair and a table ? Answer. The pattern for the chair can be chosen in twelve ways, and the pattern for the table in five ways : therefore both together can be chosen in 12 x 5 ways. Question 2. A friend shews me five Latin books, and seven Greek books, and allows me to choose one of each. What choice have I ? Answer. 5 x 7 = 35. 10 CHOICE. Question 3. If a halfpenny and a penny be tossed, in how many ways can they fall ? Answer. The halfpenny can fall in two ways, and the penny in two ways, and 2x2 = 4; therefore they can fall in four ways. The four ways, of course, are as follows : — (1) both heads. (2) both tails. (3) halfpenny head and penny tail. (4) halfpenny tail and penny head. Question 4. If two dice be thrown together, in how many ways can they fall ? Answer. The first can fall in six ways, and the second in six ways, and 6 x 6 = 36 ; therefore there are thirty-six ways in which the two dice can fall. The thirty-six ways may be represented as follows : — 1 and 1, 1 and 2, 1 and 3, 1 and 4, 1 and 5, 1 and 6, 2 and 1, 2 and 2, 2 and 3, 2 and '4, 2 and 5, 2 and 6, 3 and 1, 3 and 2, 3 and 3, 3 and 4, 3 and 5, 3 and 6, 4 and 1, 4 and 2, 4 and 3, 4 and 4, 4 and 5, 4 and 6, 5 and 1, 5 and 2, 5 and 3, 5 and 4, 5 and 5, 5 and 6, 6 and 1, 6 and 2, 6 and 3, 6 and 4, 6 and 5, 6 and 6. Question 5. In how many ways can two prizes be given to a class of ten boys, without giving both to the same boy ? Answer. The first prize can be given in ten ways, and when it is given the second can be given in nine ways, and 10 X 9 = 90 ; therefore we have the choice of ninety ways of giving the two prizes. Question 6. In how many ways can two prizes be given to a class of ten boys, it being permitted to give both to the same boy ? Answer. The first prize can be given in ten ways, and EXAMPLES OF MULTIPLICATION OF CHOICE. 11 when it is given the second can be given in ten ways ; there- fore both can be given in 10 x 10, or 100 ways. Question 7. A friend shews me five Latin books, seven Greek books, and ten French books, and allows me to choose two books, on the condition that they must not be both of the same language. Of how many selections have I choice ? Answer. I can choose a Latin and a Greek book in 5 X 7 = 35 ways, a Greek and a French book in 7 x 10 = 70 ways, a French and a Latin book in 10 x 5 = 50 ways. Therefore I have the choice of 35 + 70 + 50 = 155 ways. ^ '<- Question 8. Out of nine different pairs of gloves, in how many ways could I choose a right-hand glove and a left- hand glove, which should not form a pair ? Answer. I can choose a right-hand glove in nine ways, and a left-hand glove in nine ways, and therefore both in 81 ways, but nine of these ways would be the selection of the^nine pairs; these must^be rejected, and there remain 72 ways. \a,'' \^\ y + z vidiys ; and then the class oi y+z things can be subdivided into two classes, containing y and z things in Therefore the three classes of x, y, z things can be made in \x + y + z \y + z \x + y + z ^^^^^=^ X , or '■ |« \y + z \y \z_ \x \y \z ways (Prop. I.), which was to be proved. Corollary. We might similarly extend the reasoning if there were any more classes. Thus, the numier of ways in which v + w + x+y + z things can be divided into Jive classes, containing respectively v, w, x, y, z things, is Iv + w + x + y + z \v \w \x \y \z Example. The number of ways in which mn things can be equally divided among n persons, is \mn ■¥ (Im)". AEEANGEMENT OF THINGS NOT ALL DIFFERENT. 65 PROPOSITION VII. The number of different orders in which n things can he arranged, whereof p are all alike {of one sort), q all alike {of another sort), r all alike {of another sort) ; and the rest all different, is |p|£[r' For the operation of making this arrangement may be resolved into the several operations following : — (1) to divide the n places which have to be filled up into sets of p places, q places, r places, and n—p — q—r places respectively : (2) to place the p things all alike in the set of p places, the q things all alike in the set of q places, the r things all alike in the set of r places : (3) to arrange the remaining n—p — q — r things which are all different in the remaining set of n—p — q — r places. Now the operation (1) can be performed, by Prop. VI., |F \q\^^n—p — q—r different ways : the operation (2) can be performed in only one way: the operation (3), by Prop. III., in in—p — q — r ways. Hence, (Prop. II.) the whole operation can be performed in 1m [w — X \n —p — q — r, or \p \q\r\n- p — q — r [p |9 I*" different ways. Q. e. d. w. 5 66 CHOICE. Corollary. The same argument would apply if the number of sets of things alike were any other than three. Thus, for instance, the number of orders in which n things can he arramged, whereof p are alike, q others alike, r others alike, s others alike, and t others alike, is \p \q\r [£ |«_ Example. If there be m copies of each of n dififerent volu'meSj the number of different orders in which they can be arranged on one shelf is \mn (^r PROPOSITION VIII. Out of n different things, the number of different ways in which a selection of r things can he made, is the same as the number of different ways in which a selection of n — r things cam, be made, and is \r \n — r For either operation simply requires the n things to be divided into two sets of r and n — r things respectively, whereof one set is to be taken and the other left. Therefore (by the last proposition) whichever set be re- jected, the operation can be performed in \r In — r different ways. COMBINATIONS. 67 t The expression may be written \r\n — r n( n-l){n-2) 3.2.1 \r.(n-r){n-r-l)...B.2.1 or, dividing the numerator and denominator of the fraction by all the successive integers from 1 to n — r, n(n-l)(n-2) ... (to - r + 1) This result might have been obtained quite independently, as foUows : PROPOSITION VIII (bis). To shew that the number of ways of selecting r things out of n is n{n—l){n — 2)...{n-r + 1) Let X represent the number of ways of making a selec- tion of r things out of n things. The r things thus selected might be arranged (Prop. III.) in |r different orders. There- fore (Prop. I.) a; X Ir is the number of ways in which r things can be selected out of n things, and arranged in order. But by Prop. IV. this can be done in n(n — l)(n — 2)...{n — r + l) different ways. Therefore we have the equation a: X h- = n(n — V)(n — 2)...(n — r + l), which gives us _ n{n-l){n-2) ... (n-r + 1) ^ \ * \r the required expression. 68 CHOICE. Examples. The number of ways in which a committee of p reformers and q radicals can be selected out oi m+p reformers and n + q radicals, is m \p \n \q If there be w — 1 sets containing 2m, 3m, 4m, . . . (w — 1) m things respectively, the number of ways in which a selection can be made, consisting of m things out of each set, is |2m 1 3m |4m Wnn \mn \m \m |2m Im |3m Im |(n — l)m|m (|?w)" PROPOSITION IX. The number of ways of selecting r things out of n when repetiticyas are allowed is the same as the number of ways of selecting r things out ofn + r — 1 without repetitions. By adding all the ,n things to each selection of r out of n, we shaU get a selection of r* + m things subject to the con- dition that each thing appears at least once in every selection. Hence the number of ways required is the same as the number of ways of selecting n+r things subject to this con- dition. To make a selection oi n + r things subject to this con- dition we may take n + r blank places. Put the things into any defined order, and fill up one or more of the first places with the first thing; then one or more of the next places with the next thing, and so on. In passing from one place to the next we shall always either repeat the thing we last assigned or cha'rige to the next. Altogether in going from one end of the row to the other we make n + r — \ steps, of COMBINATIONS WITH REPETITION. 69 which n—1 must be changes, and r must be repetitions. Our only choice is to select r out of the n + r — 1 steps for our r repetitions. Therefore the choice required is the choice of r things out of w + r — 1. Q. E. D. Note. The number of ways may be written thus : l^ + ^-l _^ w(« + l)(w + 2)...(w + r--l) \n-l\r 1.2.3...r Examples. The number of dominoes in a set which goes from double blank to double n will be (w + l)(w + 2) 2 ■ ■ The number of homogeneous products of r dimensions that can be made out of n letters is n{n+l)(n + 2) ... {n + r - 1) -i- \r. Corollary. The number of combinations of r things out of n when repetitions are permitted and r>n, subject to the condition that all the n things appear in each combination, is the same as the number of ways of selecting r— n things out of r — 1 things without repetitions. Notation. We shall use the symbol P" to denote the number of permutations of n things taken r at a time, i.e., the number of ways in which an arrangement of r things can be made out of n things, as in Prop. IV. We shall use the symbol G" to denote the number of combinations of n things taken r at a time, i.e., the number of ways of selecting r things out of n things, as in Prop. VIII. And we shall use the symbol B^ to denote the number of 70 CHOICE. ways of selecting r things out of n when repetitions are allowed as in Prop. IX. Remember that p» = re(«-l) (?i-2)... {n-r + i), _ n{n-l)(n-2)...{ji-r + 1) \.'2,.Z...r P„_ w(w + 1) (w + 2) ...(w + r-1) ^'~ 1.2.3...r- Also r: = G",'-'-\ \n And c: = c:_=- \r \n — r It is convenient to consider that the full symbol for the number of combinations oi x + y things taken x together or y together, or for the number of ways of dividing x + y things into a set of x and a set of y things is G'*^. But as the affix in this symbol is always the sum of the two suffixes, it is only necessary to write two of the three. Thus we may write indiflferently Cr, C'l^", or G^„ and again a:, o» , or a „ ,. vJ n—T T, n — T It is convenient to notice that C'to+I. „-1 ^ + 1 Cm; nJ and, therefore, G G TO' n ^TO+1' n-i m — n+1 TO+1 c„,, _ TO - 71 + 1 ' TO+n+l "'+'' and, as a particular case, - 1 r -' n + l " 1' m ~ 20„,„ 2^«+l' n+r Also that ^n' »-l = iO^: > n" WHOLE NUMBER OF SELECTIONS. 71 PEOPOSITION X. Out of n different things, when each may he repeated as often as we please, the number of ways in which an arrange- ment of r things can be made is rJ. For the first place can be filled up in n ways, and when it is filled up the second place can also be filled up in n ways (since we are not now precluded from repeating the selection already made) ; and so the third can be filled up in n ways, and so on, for all the r places. Hence (Prop. II., Cor.) all the r places can be filled up, or the whole arrangement can be made, in w' different ways. Q. E. D. PROPOSITION XI. The whole number of ways in which a selection can he made out of n different things is 2" — 1. For each thing can be either taken or left ; that is, it can be disposed of in two ways. Therefore (Prop. II., Cor.) all the things can be disposed of in 2" ways. This, however, includes the case in which all the things are rejected, which is inadmissible ; therefore the whole number of admissible ways is 2" — 1. Q. E. D. CoEOLLAKT. Since the total number of selections must be made up of the combinations taken one at a time, two together, three together, &c. and n together, we must have 0^+C?+CS+... + C^ = 2»-l, 72 CHOICE. PKOPOSITION XII. The whole number of ways in which a selection can be made out of p + q + r + '-'-/+ ...+/=(« + y)". Suppose we have n things to give to x men and y women without any restriction as to the number given to each. PROOF OF THE BINOMIAL THEOREM. 73 Since there are « + y persons, we may give the things in {oB + yf ways. But we may classify these ways according to the number of things given to men and women respectively. If we give n — r things to men and r things to women, we may select the r things for the women in 0^ ways, and then assign the remaining n — r things to the men, and the selected r to the women, in af^'y" ways. Thus we get O^x"'" y" different dis- positions of the things. And r may have all values from to n. Giving r all these values in succession we get « + 1 sets of ways of disposing of the things, which together must be the total number of ways (x + ^)" which we obtained at first. Thus we have a;" + Glx'-'^y + Glap-^f + ... + G^.aP-^if + ... + j/" = (a; + y)". Corollary. The proposition must also be true when X and y are not positive integers. For the result of the continued multiplication of a; + y by itself must be the same in form whatever x and y may represent. Therefore always, if m be a positive integer, (^ + 2/)" = ^" + f ^-'y + '^^^«'""2/' + • • ■ This formula for the expansion of (x + y)" is called the Binomial Theorem. PROPOSITION XIV. If there he N sets of lettei~s^, and if out of r assigned letters a, j8, 7 ... each letter such as a. occur in N^ sets; each 1 It should be observed that the sets may be parcels or groups, selections or arrangements, with or without repetitions, and they will not neceesarily all contain the same number of letters. 74 CHOICE. combination of two letters {such as a, /3) in N^ sets; each com- bination of three letters (such as a, /3, j) occur in N^ sets; a/nd- so on; and finally all the r letters occur in N^ sets, then the number of sets free from all the letters will be the coefficients following the law of the Binomial Theorem. The number of sets free from a is N — N^ Among these, jS is found in If^ — N^ .'. (by Subtraction) the number of sets free from a or /3 is N—^N^ + N^ Among these, 7 is found in N^ — 2N.^ + N^ (for, if we had introduced from the be- ginning the condition that 7 be present we should have had throughout N^, N^, iVj for N, N^, N^, respectively.) .". (by Subtraction) the number of sets free from a, /3, or 7 is N -^N^-\-ZN^- N^ Among these S is found in N,-^N^ + ^N^- N^ (for if we had introduced from the be- ginning the condition that S be always present we should have had N^, N^, N^ ... for N, N^, N^... respectively.) .'. (by Subtraction) the number of sets free from a, /8, 7 or S, is iV-4iV, -1- QN^-'^N^ + N^ And so on. Now we may observe that the successive operations per- formed at the right-hand side of the page are identical with the operation of constantly multiplying by 1 — i\r except that we have N,N^, N^ N^... in place of 1, N, N'\ N\.. respec- REGUEEENCE OF CONDITIONS. 75 tively. Therefore the coefficients must be the same as in the involution of 1 — N, i.e. they must follow the law of the Binomial Theorom. Therefore we have (after r operations), the number of sets free from any of the r letters, viz.: ,_ r T,r r(r — 1) „ r(r — 1) (r — 2) ,^ „ ,^ -^~ 1 -^^ + 1.2 ^'~ 1.2.3 -^s + &c. ...±N^. Q. E. D. Corollary I. If all the sets must contain some of the letters the expression just found must be zero, and therefore we find -N in terms of N^, N^, N^, &c. ; thus ,-. r ,-r r(r— 1) ,^ r (r — 1) (r- — 2) ,^ „ _ „ ■^^T-^'" T72"^^-^''+ 1.2.3 ^s- &<>■■■■ +K- Corollary II. Suppose the N sets of letters are the combinations of m letters taken w at a time. The number free from any of r assigned letters will be (7""''. And we have N = C™ N^ = GZ:l, N, = GZll and so on. Therefore O™-' =GZ-Gl (7™V + Gl (7™t - &c. + GZ:;. In the particular case when w = m — r we get 1 = 6^r - CT'' Gl + CT'' Gl - &c. to (r + 1) terms. CoEOLLARY III. Our reasoning being perfectly general we may enunciate our proposition more generally as follows. If there he N events w operations and if {out of r possible conditions a, /8, 7,...) every one condition {such as a) be ful- filled in iVj of the events ; and enery combination of two simul- taneous conditions {such as a, /8) be fulfilled in N^ of the events ; and every combination of three simultaneous conditions 76 CHOICE. (such as a, y8, 7; he fulfilled in JV, of the events ; and so on ; and finally all the r conditions he simultaneously fulfilled im N^of the events ; then the number of events free from all these conditions is N-tN^ + '-^^. - - (7\)(7^) i^, + &c.... ± if,. Example. A symmetrical function of a, h, c, d has one term without any of these letters; a is found. in 9 terms, ah in 5 terms, ahc in 3 terms, and abed in 2 terms. How many terms are there altogether ? Here we have JV; = 9,N^ = 5, N^ =3,^^= 2, and 1 = if - 4if, + 6if,- 4if, + if^ or l=Jf-36 + 30-12 + 2; .■.N=17. N.B. The seventeen terms may be such as l + a^+h^ + c^ + d'' + hcd + acd + ahd+ ahc + ahcd + a^¥ + r>n) or = GT-: + Gi G^::^, + gi or„%, + &c. to (w + 1) terms, the last term being (7™"". 78 CHOICE. 3. Again the selections of r letters out of m may be obtained by taking (1) those which contain a, (2) all of the rest which contain /3, (3) all of the rest which contain 7, and so on until only r — 1 letters remain. Hence (7r = O'^ll + 0:^1' + G'^-.l + tom-r+l terms, the last term being O^lJ or unity. Writing n ior m — r and reversing the series, we get 1 + Gi, + c;Ji + 0:-^, + ... + G:;:^-' = c:^ or, (which is the same thing) i + g;-\- C'r-' + cr' + ...+0;+^-'=C'r. 4. When repetitions are allowed (Prop. IX.) the selections of r letters out of m letters may be classified according as a is used 0, 1, 2, 3... or r times. Hence J?™ = ^-^ + ^_-i + E™-/ +. ..+ B^-' + 1, or 0^-^= cr^-=+o:!+'-'' + (7::^-*+...+ cr' + i. Writing w + 2 for m, and reversing the series, 1 + or' + g;^ + Gf +. . .+ a:r^ = CT^\ the same result as in Art. 3, with w + 1 for r, and r for n. 5. By Art. 3 or 4 we have 1 + G^^+ 01^^+0;^+. ..ton terms = G'„% or 1 + ar' + G;-^ + Cr' +. . .to n terms = C,f • Multiplying throughout by Ir we get P:: + P^' + P:""' +• ■ -to M terms = ^^ . r + I SUMMATION OP SERIES. 79 Giving r the values 1, 2, 3... in succession, we get the following series each summed to n terms, 1+2 + 3+.. . =in(n+l), 1.2 + 2. 3 + 3.4+... =Jw(w+l)(« + 2), 1.2. 3 +2. 3. 4 +3. 4.5+... =in(n + 1) (n+ 2) (w + 3). 6. Suppose we have to choose n persons out of n men and n women. The number of selections is 0^. We may classify these selections according to the number of men in each, observing that the number containing r men and n — r women is C" . C"_ or (C")^ Thus we find {.Off + {CXf^■{Gtf +...+ ((?:)= = (7^- 7. In the last article, if there had been x men and y women, x and y being each greater than n, we should have found Gl + GU G{ + OU Gl+...+ Gl= GT- 8. The symbol S placed before a function of a variable {x suppose) will denote the sum of the different values which the function takes, when x has successive integral values between stated limits. The limiting values of x are generally marked above and below the S. Thus the results of the foregoing articles may be concisely written t-{GT-']=G^. Vn-1 f p»'+*l -*- -pr-\^ —0 I »• J ~ ^ + 1 • -^'■+1' 80 CHOICE. 9. To shew that lm + n— I m— 1 n—1 By Prop. IX. the number of combinations of n things out of TO, repetitions being allowed, is G'^'\ We may classify these according as 1, 2, 3... or w of the to things be used. If « things be used, they can be selected in C" ways; and a combination of n things (using all the x things) can be made in Cl:l ways (Prop. IX. Cor.)'.. Therefore the whole number of combinations using exactly aJthingsisOrO^i Giving a; all values from 1 to w we shall get all the com- binations. Therefore But G:-l = -Gl., = -G:. Hence |m + w — 1 %l{xG':Gl]=nGT""' = Ito — 1 |?i — 1 ' or TO w to(to — 1) n{n — l) T'l"'' iT2 • 1.2 „ TO (to -1) (to -2] n (n - \) {n - 2) . + ^ 17273 ■ 17273 +*'■ to n terms if w < to | |to + w — 1 to m terms if to < w ^ I |to + w — 1 ■ i Ito - 1 Iw — 1 10. If there be in a bag to black balls and n white balls they can be drawn out, one by one, in (7"'^ orders. Let us classify these orders according to the number of white balls drawn before the A'" black ball. SUMMATION OF SERIES. 81 If X white balls be drawn before the k^ black one the x + k—1 balls can be drawn in 0***"' orders and the subse- quent balls in G'^'"-' orders. And X may have any value from to n. Therefore V« {rlx+k-\ flm+n-k-x') _^ rim+n ({x + k — llm + n — k — x^ \m + n or -S?j ' ' W=- \^\x\k—\\n — x\m — k) \m\n Therefore (W]x + h--l\m + n — k — x'\ {m + n) \m — k\k — \ ^»{oi?}= ~kGf- Writing pioxm + n—1 and g' for A; — 1, •^0 n n{n — V) !_ 1 1.2 1 _j) + l 1 11. In the last article, we had writing p for m + w — Z; we get or Cr; + f C-1 + ^^ Gl-.\ + ... + ar- = or*. 12. In Prop. XIII. we proved the Binomial Theorem for a positive integral exponent. We must refer to Treatises on Algebra for the proof that it is true whatever the exponent be ; so that always w. 6 82 CHOICE. /I . \n -I w wCn — 1") „ n(n — l)(n — 2)^ „ {l + xy = l + ^x+ \2 ''+ 1.2.3 ^^ +'^c. the series stopping at the (m + l)th term when w is a positive' integer but going on ad infinitum when n is fractional or negative. The following are some important examples, which we shall have to refer to hereafter. > (1 -4«)-*= 1 + Glx + G^a? + Clx^ + &c. (1 - 4«;)* = 1 - 2x - 2CJ I' - 20^ ^ - 2C1 |V &c. 13. Since the product of (1 — 4a;)* and- (1 — 4*)"* is unity, the coefficient of «"■" in the product of their expan- sions must be zero. Therefore c=;::^-2|(?+ic^.ic?+ jc^:-c^+ ... +^^ (p:| =0, __ 1 ron-w _ re»H-i 2 n+l n ' 14. From the expansion of (1 — 4a;)~*, squaring, we obtain (1 -4a;)-^= {1 + Glx+C!\x^+CP,x^+...]\ But by the Binomial Theorem (1 - 4a;)-'.= 1 + 4a; + (4a;)' + (4a;)' + . . . Equating the coefficients of a;" in these two expansions we get , C+C?C-r+C'^''+ ^^ a^'+ ... i^-iy =x^ +y+ |«,"+1|-V+ f ^"+ ^^ «,"+... X CHAPTER III. DISTRIBUTIONS. Most of the questions of Permutations and Combinations which we have considered have involved the division of a given series of things into two parts, one part to be chosen, and the other rejected. The theorem expressed arithmeti- cally in Rule VI. (page 27), and algebraically in Proposition VI. (page 64), is the only one in which we have contem- plated distribution into more than two classes. But as the number of things to be given to each class was in the terms of that theorem assigned, the problem was reduced to a case of successive selection, and was therefore classed with other questions of combinations. But when the number of ele- ments to be distributed to each several class is unassigned, and left to the exercise of a further choice, the character of the problem is very much altered, and the problem ranks among a large variety which we class together as problems of Distribution. Distribution is the separation of a series of elements into a series of classes. The great variety that exists among problems of distribution may be mostly traced to five prin- cipal elements of distinction, which it will be well to con- sider in detail before enunciating the propositions on which the solution of the problems will depend. 86 CHOICE. I. The things to be distributed may be different or indif- ferent. The number of ways of distributing five gifts among three recipients, will depend upon whether the gifts are all alike or various. If they are all alike (or, though unlike, yet indifferent as far as the purposes of the problem are concerned), the only questions will be (i) whether we shall divide them into sets of 2, 2, 1 or 3, 1, 1, and (ii) how we shall assign the three sets to the three individuals. If on the contrary the five gifts are essentially different, as a, b, c, d, e, then they may be divided into sets of 2, 2, 1 in 15 ways, and into sets of 3, 1, 1 in 10 ways, and then we shall have to assign the three sets which are in this case all essen- tially different (because their component elements are so), to the three individuals. In the first case, the sets could be formed in 2 ways, and when formed in either way they could be assigned in 3 ways, thus giving a complete choice of 6 distributions. In the second case, the sets could be formed in 25 ways, and when formed in any way they could be assigned in 6 ways, thus giving a complete choice of 150 distributions. II. The classes into which the things are to be distributed may be themselves different or indifferent. We here use the adjectives different or indifferent to qualify the abstract classes regarded as ends or objects to which the articles are to .be devoted, without any reference to a posteriori differ- ences existing merely in differences of distribution into the classes. , Where five gifts were to be distributed to three reci- pients, the distinct personality of the three recipients made the classes characteristically different, quite apart from the consideration of the differences of the elements which com- posed them. But if we had only to wrap up five books in GROUPS AND PARCELS. '87 three different parcels, and no difference of. destination were assigned to the parcels, we should speak of the parcels as indifferent. The problem would be simply to divide the five things into three sets, without assigning to the sets any particular order. The distribution could be made in 2 ways if the things themselves were indifferent, and in 25 ways if they were different. III. The order of the things in the classes may he differ- ent or indifferent, that is, the classes may contain permuta- tions or combinations. Of course this distinction can only arise when the things themselves are different, for we cannot recognise any order among indifferent elements. We shall avoid confusion by distinguishing arranged and unarranged classes respectively as groups and parcels. If three men are to divide a set of books amongst them, it is a case of divisioiji into parcels, for it does not matter in what order or arrange- ment any particular man gets his books. But if a series of flags are to be exhibited as a: signal on three masts, it is a case of division into groups, for every different arrangement of the same flags on any particular mast would constitute a different signal. IV. It may or may not be permissible to leave some of the possible classes empty. It will entirely depend upon the circumstances out of which the problem arises, whether it shall be necessary to place at least one element in every class, or whether some of them may be left vacant ; in fact, whether the number of classes named in the problem is named as a limit not to be transgressed, or as a condition to be exactly fulfilled. If we are to distribute five gifts to three recipients, it will probably be expected, and unless otherwise expressly stated it will be implied, that no one 88 CHOICE. goes away empty. But if it be asked how many signals can be displayed by the aid of five flags on a three-masted ship, it will be necessary to include the signals which could be given by placing all the flags on one mast, or on two masts. V. It may or may not he permissible to leave some of the distributable things undistributed. This will be illustrated by a comparison of the Propositions XV. and XVIII. below. The following table in which the cases considered in the present chapter are classified, will be useful as an index to the propositions. Nature of the distribution. Whether blank lots are allowed. Whether all the things must be distributed. PROP. Distribution of different ■ things into ' different groups different „ indifferent „ different „ different „ indifferent „ Tes No No Yes No No Yes Yes Yes No No No XV. XVI. XVII. XVIII. XTX XX Distribution of different things into different parcels Yes No Yes Yes XXI. xxn. Distribution of different things into indifferent parcels No Yes Yes Yes xxni. XXTV. Distribution of indifferent things into different parcels No Yes the lots ) limited j Yes Yes Yes Yes XXV. XXVI. XXVII. XXVIII. Distribution of indifferent things into indifferent parcels No Yes XXIX. XXX. DISTRIBUTION INTO GROUPS. 89 PROPOSITION XV. The number of ways in which n different things can he arranged in r different groups {blank groups being admissihle) is r{r + l)(r + 2) ... {r + n-1). For we shall obtain the arrangements by taking the // different things and putting with them r — 1 indifferent points of partition, and arranging these n + r—1 things in all possible orders. Since i — 1 of them are alike the number of arrangements is by Prop. VII. \n + r—l -5- Ir* — 1, or r (r- + 1) (j- + 2) ... (r + « — 1). q.e.D. Example. The number of ways of arranging n flags on r masts when all the flags must be displayed, but all the masts need not be used, is r(r + l) (r + 2) ... {r + n—1). PROPOSITION XVI. The number of ways in which n different things can be arranged in r different groups (none being blank) is In In^—l \n GUI or ■ . \n — r\r — \ For we shall obtain the arrangements by (1) arrangiDg the n things in all possible orders, and (2) inserting r — 1 points of partition in a selection of r — I out of the n — 1 in- tervals between them. 90 CHOICE. The arrangement (1) can be made in |m ways, and the selection (2) in Cl'_[ ways. Therefore our distribution can be made in ]n GlZ\ or \n |« — 1 -=- |m - r | r — 1 different ways. Q. E. D. Example. The number of ways in which n flags can be arranged on r masts so that every mast has at least one flag is \n \n — 1 -i- \n — r \r — 1. PROPOSITION XVII. The number of ways in which n different things can be arranged in r indifferent groups {none being blank) is \n \n In — 1 ■--cr.\ or '- ' — -. \r \n — r\r\i — 1 . For it is plain that for any one distribution in this case we must have I n arrangements when the groups are not in- different. Hence we have only to divide the result of the last proposition by Ir; which gives the required expression. PROPOSITION XVIII. Out of n different things the total number of ways in which r different groups can be made {blank groups being admis- sible) is the coefficient ofx" in the expansion of ]ne' DISTEIBUTION INTO GROUPS. 91 For if we use x things we may select them in C" ways, and then distribute them (by Prop. XV.) in r (r + 1) (r + 2) ... (r + a; — 1) ways. Tlius we get CZ.r{r+\) (r + 2) ...{r + x-\) distributions. And (including the case when all the groups are blank) x may have every value from to ji. Therefore the total number of distributions is 1 ^ r ^ r{r+r) ^ r-(r + l)(r+2) ^ ^^ \rh |1 \n-l |2 |w-2 [3 |w - 3 to 71+ 1 terms, which is the coeflScient of x" in the. product of the two series n n + — +— + — +&C. - I [1 [2 [3 and l + ^. + ^(^ + ^) .'- + ^(^+^)(^+^) .'+&c., and these are respectively the expansions of \n ef and (1 — a;)"'. Hence the total number of distributions is the coefficient of x" in the expansion of [we' (rr<- Examples. This proposition will give the total nuinber of signals which can be made by displaying arrangements out of n flags upon r masts, when one is not required to use either all the masts or all the flags. Writing n = 12 and r = 8, we should have the number of ways in which a lady 92 CHOICE. with 12 rings could wear some or all of them on her 8 fingers. PROPOSITION XIX. Out of n different things the total number of ways in which r different groups can be made (without blanks) is the coeffi- cient ofx"'' in the expansion of \n e" (1-^r For if we use x things, we may select them in In -=- 1« — a; \x ways, and then distribute them (by Prop. XVI.) in la; la; — 1 -H la; — r [r — 1 ways. \n |a;— 1 Thus we get \n— X \x — r \r ■ distributions. And x may have every value from r to n in- clusive. Therefore the total number of distributions is I f 1 , r r(r + l) r(r + l)(r + 2) I— [ \n — r |1 \n — r — l \2 n — r—2 [3 \n — r — S + &c. to n — r + 1 terms > , which (as in the last proposition) is the coefficient of a;""' in the expansion of Ine' L . Q. E. D. (1-xy DISTRIBUTIONS. 93 PROPOSITION XX. Out of n different things the total number of ways in which r indifferent groups can be made (without blanks) is the coeffi- cient of aS" in the expansion of Ine" |r(l -«)■"■ This follows from Prop. XIX., as Prop. XVII. from Prop. XVI. Example. This proposition wDl give the number of ways of forming r trains at a railway station at which there are n carriages. The foregoing propositions have related to distribution into groups, the remainder of this chapter will be devoted to distribution into parcels. PROPOSITION XXI. The number of ways in which n different things can be distributed into r different parcels is r", when blank lots are admissible. For each of the n different things can be assigned to any one of the r parcels without thought of how the others are disposed of. Hence the n things can be successively dis- posed of in r ways each, and therefore (Prop. II.) all can be disposed of in r" different ways. Q. E. D. PROPOSITION XXII. The number of ways in which n different things can be distributed into r different parcels (without blank lots) is |n times the coefficient ofx" in the expansion of (e* — 1)'. 94 CHOICE. For by the last proposition, the number of distributions ■when blanks are allowed is r". And the number of distribu- tions subject to the condition that an assigned parcel is blank is (r — 1)". And the number of distributions subject to the condition that two assigned parcels are blank is (r — 2)" and so on. Hence, by Prop. XIV. Cor. 3, the number of distribu- tions in which none of the r parcels are blank is -f&c....+ j.2" + l, that is (Excursus, Art. 17), \n times the coeflScient of ic" in the expansion of {f — 1)'. Q. E. D. Examples. The number of ways in which five different commissions can be executed by three messengers is 3° or 243. But if no one of the messengers is to be unemployed, the number of ways will be 15 times the coefficient of x^ in the expansion of {f — 1)' ; that is,. (Excursus, Art. 18) 5 2" X [5, or 150. There are 8" or 8589934592 ways of giving 11 different gifts to 8 boys. But if no boy is to be without a gift the number of ways is [11 times the coefficient of a;" in the ex- pansion of {^- Vf, that is 12 X |11 or 479001600. PROPOSITION XXIII. The number of r-partitions of n different things, i.e. the number of ways in which n different things can- be distributed DISTEIBUTION INTO PAECELS. 95 into r indifferent parcels, with no blank lots, is \n times the coefficient of of in the expansion of t This follows from Prop. XXII., as Prop. XVII. from Prop. XVI., or Prop. XX. from Prop. XIX. Example. To divide the letters a, b, c, d, e into three parcels. The number of ways will be 15 times the coefficient of iB^ in the expansion of {e° — 1)' h- 13 ; that is (Excursus, Art. 18), 5 . 4 X I = 25. The twenty-five partitions are easily seen to be ten such as ahc, d, e, and fifteen such as ab, cd, e. PROPOSITION XXIV. The total number of ways in which n different things can be distributed into 1, 2, 3... or n indifferent parcels is \n times the coefficient of «" in the expansion of — . e Let N, denote the number of r-partitions of n different things; then e* — 1 N^=\n times the coefficient of a;" in , |1 N^ (^^-1)' and so on. 96 CHOICE. Therefore by addition is equal to In times the coefficient of x" in the expansion of -^_1 (^_1)« (^^_1)»^ \l ^ ^ this last series being carried to infinity if we please, since the terms beyond the n^ do not involve db", and therefore the inclusion of them will not affect the coefficient of ss". But this series is the expansion of e'"'"^ — 1, and the co- efficient of x" therein is the same as in the expansion of gea:-i Qj. ^ ^ g Hence N^ + JSr^ + N^ + .-. + N^ is equal to in times the coefficient of «" in the expansion of — . Q. E. D. ■e PROPOSITION XXV. The number of ways in which n indifferent things can he distributed into r different parcels (blank lots being inadmissi- ble) is the number of combinations of n—1 things taken r—1 at a time. For we may perform the operation by placing the n things in a row, then placing r — 1 points of partition amongst them, and assigning the r parts thus created, in order, to the r parcels in order. Hence the number of ways is the number of ways of placing r — 1 points of partition in a selection out of w — 1 intervals. Therefore it is the same as the number of combi- nations of w — 1 things taken r — 1 at a time. Q. E. D. DISTRIBUTION INTO PAECELS. 97 PROPOSITION XXVI. The number of ways in which n indifferent things can he ■distributed into r different parcels (blank lots being ad/nissi- hle) is the number of combinations of n + r — 1 things' taken r—1 at a time. For the distribution of n things, when blank lots are admissible, is the same as the distribution oi n+r things when they are not admissible, since in the latter case we have to place one thing in each of the r parcels, and then' to distribute the remainder as if blank lots were' admissible. Hence, writing ji + r for w in the result of Proposition XXV., we obtain the number required. Examples. Twenty shots are to be fired ; the work can be distributed among four guns in ' ' or 969 ways, without lea'ving any gun unemployed. Or, neglecting this 23 22 . 21 restriction, the work can be done in „ ' or 1771 ways. Again, five partners in a game require to score 36 to win. The number of ways in which they may share "this score (not all necessarily contributing) is 40 ■ 39 . 38 . 37 1.2.3.4 ' or 91390 di£ferent ways. Again, in how many ways can five oranges be distributed amongst seven boys ? Evidently two or more of them will get none. The answer is 462, viz., ll.t«i.^.X.7.X 1 .2..'a.-^.'S..'«i ■ w. 7 98 CHOICE. PROPOSITION XXVII. The number of ways in which n indifferent things can be distributed into r different parcels, no parcel to contain less than q things, is the number of combinations ofn — 1 —r(q — l) things taken r — 1 at a tim£. For if we first place q things in each of the r parcels we shall have n — qr things left, and it will only remain to dis- tribute them among the same r parcels according to Proposi- tion XXVI., which shews that the number of ways of making the distribution is the number of combinations oin — qr+r — \ things taken r — 1 at a time, q.e.d. PROPOSITION XXVIII. The number of ways in which n indifferent things can be distributed into r different parcels, no parcel to contain less than q things, nor more than q + z — 1 things, is the coefficient of x"'" in the expansion of (^O- For if we multiply together r factors, each represented by a;» + a!«+'-l- + x^-\ we shall have in our result a term x" for every way in which we can make up n by the addition of one index qoi q+\ or q-\i2 (k &c,, ox q + z—1 from each of the r factors. Hence we shall hai/e as" as many times as there are ways of dis- tributing n into r different parts, no part less than q nor DISTRIBUTIONS. 99 greater than q + z — 1. Therefore the number of ways of so distributing n is the coe£Scient of «" in the expansion of orof ar(l+ai + x'+...af-^y, which is the coefficient of x"~^ in the expansion of (l+x + a!'+...+af-y, orof (^). Q.E.D. Example. The number of ways in which four persons, each throwing a single die once, can score 17 amongst them is the coefficient of x"~* in the expansion of (^y Now (1 - xj = 1 _ 4a;' + 6a;" - &c. (1 - a;)-* = ^ |l . 2 . 3 + 2 . 3 . 4a; + 3 . 4 . 53!" + . . .1 . And the coefficient of x" in the product = i ■ 14 . 15 . 16 - 4 . 8 . 9 . 10 + 6 . 2 . 3 . 4i = 104. The following notation will be sometimes convenient. Let 71 be a number not divisible by r. Then - is not an integer, it therefore lies between two consecutive integers. Let — denote the lesser of these and let — denote the r r greater. Then we shall have r r *100 CHOICE. Also let —^^ denote. the dnteger nearest to -, so that n + m Tf n — -^ = — or — , -r r r according as the division of n by r- leaves a remainder greater or less than ^ . . . ■ ; It must only be noted that if the remainder is equal to s the expression — = would be ambiguous, and we must use either — or as the case requires. r r Further, let it be understood that if n happen to be divisible by r, each of the expressions , — , —^ must be interpreted to mean - . Then, generally, — is the greatest integer in - , — = is the nearest integer to - , — is the least integer not less than^ . r ^ r PROPOSITION XXIX. To 'find the number of ways in iuhich n indifferent things can be distributed into r indifferent parcels {no blank lots). Or To find the number of different r-partitions of n. ii-PARTITIONS OF N. 101 Let 11" denote the number of r-partitions of n, or the number of ways of distributing n indifferent elements into r indifferent parcels. Suppose that in any distribution, a; + 1 is the smallest number found in any parcel. Then setting aside a parcel which contains a; + 1 all the other parcels contain not less than x + 1, and therefore more than x. If we place x in each of these r —1 parcels, the distribution can then be completed by distributing the remaining n — l—rx things among the same r—1 parcels, and this can be done in jjn-i-r. yfayg_ jjj^ ^jjjg y^gj -yyg shaU obtalu all the distribu- tions, by giving a; successively all its possible values. But since a; + 1 is the smallest number found in any parcel, x+1 cannot be greater than the greatest integer in -, or . Then x must have all values from to 1, and therefore r np = W^zl + n^-^-' + Tl^ll-^ + ... to — terms: Now it is plain that n" = 1 for all values of n. Hence H? = Hr ' + Hr ' + U^' + . . . to ^ terms = ^ . Again HJ = H"-' + Hr' + HJ"' + ... to ^ terms « — 1— m — 4— ,w — 7— ,w — , = -^— + -^- + —2- + - to-3- t^'^"^^- The summation will depend upon the form of n ; thus, Iin = 6q then IIj = j^ 102 CHOICE. n = 6q±2 n» = ^ n = 6q + S n^ = - 12 Therefore IIJ is always the integer nearest to rs , or "» 12 • Again, HJ = Ilr' + H^' + n^"" + . . . to ^ terms . (n-iy±An-5y± (n-9y± ^^fLZ terms 12 ^ 12 + 12 +"-™ 4, *^"^s. The summation will depend upon the form of n ; thus. « = 12g' then n:^ = n = 123 + 1 u:= w = 12gr + 2 n: = w = 122 + 3 nr = w = 123 + 4 u:= n = 12g + 5 ii:= n = 122- + 6 n» = n = 12^ - 5 nr= 144 k' + 3?i=-9w + 5 144 n' + Sn" - 20 144 n' + Sn'-dn- ■27 144 w'+3n'' + 32 144 n^ + Sn'-dn- ■11 144 n.' + Sn' - 36 144 to' + Sm' - 9w + 5 144 J8-PABTITI0NS OF N. 103 n = 12q -4 "^ ~ 144 n = 12q-S n„ rv' + Bn'-9n-27 144 n = 12^ - 2 »i' + 3ri»-4 * ~ 144 n = 12q-l TT„ m' + 371" -9m -11 "^ ~ 144 Therefore, = — ZTTT — when n is even : n' + 3n"-9n+ , . ,, = jj^ = when n is odd. By a like process we may deduce successively 11", IIj, &c., and thus we may find 11" for any values of n and r, although we cannot write down a general expression for 11" in any simple terms. Examples. There are twelve 3-partitions of 12, viz. 1> 1, 1, 10 1, 3, 8 1, 5, 6 2, 3, 7 2, 5, 5 3, 4, 2, 9 1, 4, 7 2, 2, 8 2, 4, 6 3, 3, 6 4, 4, There are fifteen 4-partitions of 12, viz. 5 4 1, 1, 1, 1, 1, 9 1, 1, 4, 6 1, 2, 3, 6 1, 3, 4, 4 2, 2, 4, 1, 2, 8 1, 1, 5, 5 1, 2, 4, 5 2, 2, 2, 6 2, 3, 3, 1, 3, 7 1, 2, 2, 7 1> 3, 3, 5 2, 2, 3, 5 3, 3, 3, 4 4 3 The number of 4-partitions of 13 is the integer nearest to (w' + 3«* — 9»i)-^144 when w = 13. Therefore there are 18 partitions, viz. 1, 2, 5, 5 2, 2, 2, 7 1, 3, 3, 6 2, 2, 3, 6 1, 3, 4, 5 2, 2, 4, 5 1, 4, 4, 4 2, 3, 3, 5 1, 1, 1, 10 1, 1, 2, 9 1, 1, 3, 8 1, 1, 4, 7 1, 1, 5, 6 1, 2, 2, 8 1, 2, 3, 7 1, 2, 4, 6 2, 3, 4, 4 3, 3, 3, 4 104 CHOICE. It appears from what precedes that the number of r-partitions of any number is easily calculated if we have a register of the (r — l)-partitions of lower numbers. There is no difficulty therefore in constructing a table of the number of r-partitions of n for a series of consecutive values of r and n, commencing with unity. Such a table is given on page 106. It is however most readily constructed by the aid of the following theorem. PROPOSITION XXX. To prove that the nwrriber of r-partitions of n exceeds the number of {r — l)-partitions of n — 1, by the nv/mber of r- partitions of n—r. To make every r-partition of n, we must first take r of the elements and place one of them in each of the proposed parcels. We have then to distribute the reiriaining n — r elements into 1, 2, 3 ... or more of the parcels. Hence n^ = UT + nr + nr' + . . . + n;rr + n^-'. Now write n^l for n, and r — 1 for r, then n::i = nr + nr + n^ + . . . + n^x therefore by subtraction n;: - n^ij = nr. CoEOLLAET. lin — r 5 , IT" = 11""^ We proceed to shew how the table of partitions given on page 106 is most readily constructed, or how it can be most readily extended at will. /J-PAETITIONS OF N. 105 We commence by observing that for all values of n n"=i 11''= 1 n" =1- therefore we may fill, the highest row and the two lowest diagonal rows with unity. Also, since if a; > «, II^ = 0, we may regard all the blank spaces below the lowest diagonal as fiUed with zero. This makes a sufficient commencement of the table, though it may be most convenient to insert at once the row of 2-partitions, which we know to be 1, 1, 2, 2, 3, 3, &c. since ih - 2 ■ We have now to see how the table when once commenced and carried to any extent may be further extended. Take any vertical column, as for instance, the column under the head 7. It reads thus 1, 3, 4, 3, 2, 1, 1, 0, 0, 0, &c. Ad4 up the sum of this series first to one term, then add another terrn, and another, and so one ; we get 1, 4, 8, 11, 13, 14, 15, 15, 15, 15, &c. But by the proposition these sums must be respectively III, n', Hf, and so on, ad infinitum. That is, they consti- tute a diagonal of our table commencing under the head 8, and extending indefinitely downwards to the right. In this way it is quite easy to write out the table to any extent, proceeding by successive diagonals, each entry in the table being found by the simple addition of two numbers. 106 CHOICE. C9 CO CO cq C3 cq i-i 00 l-H c- l-H CO iH in l-H 1-1 CO ^ \p\ ^^ ^^ ^ ^* ^ ^ ^^ ^<^ ^ r^ lo -vM en rsi ^-i rH tH C5 ^ (M lO tH O lO CO W iH I> U3 CO (M rH THOe0"*-*O(NO r-l CO CO 00 05 00 t* »H t- US CO (M iH rH iH t- UD CO CT W i-l t-D-000500»OCO«i-ti-l »CCIU5CO-000>0 I— li-Hi— li— li— !r— 4i— li— li— li— ICQ CHAPTER IV. DEKANGEMENTS. When we place a series of elements in a particular order we are said to arrange them. But if they have been already arranged, or if they have a proper order of their own, and we place them in other order, we are said to derange them. Thus derangement implies a previous arrangement in which each element had its own proper place, either naturally belonging to it or arbitrarily assigned to it. If we begin with unity and multiply successively by 1, 2, 3 ... as if we were going to form Iw, but let every even product be increased by unity and every odd product be diminished by unity, we shall obtain a series of numbers which we shall denote by the symbols || 1, || 2, || 3, &c. Thus multiply by 1 and subtract 1 : ' we get || 1 = 0, then multiply by 2 and add 1 : we get || 2 = 1, then multiply by 3 and subtract 1 : we get || 3 = 2, then multiply by 4 amd add 1 : we get || 4 = 9, and so on. Or, more fully, 112 = 2.1-2+1, . 113 = 3.2.1-3.2 + 3-1, II 4 = 4. 8. 2. 1-4.3.2 + 4.3-4+1. 108 CHOICE, and finally lbL=l!^|l-jI + g-[3 + -±l^, As In is called faotorial-n it is convenient to read \\ri as svb-factorial-n. The Values of the first twelve factorials and sub-factorials, and their ratios, are given on page 114. Since by definition ||rt + l = {n+l)\\n±l, and \\n=n \\n-l + 1, we have by addition ||w+l + |[w = (w + l) \\n + n \\n-l , or ||w+l = n (^\\n + \\.n - l y This equation expresses the law connecting each factorial with the two preceding ones ; and it is curious to note that the same law holds good for factorials and for sub-factorials, since \n'+ 1 = n (\n + \n — ly Comparing the series for \\n with the expansion of e"' (page 83) we deduce \n upper sign when n is odd, — ~'e~ ' lower sign when n is even, where fi^ is a small firaction lying between and ^ . •'^ w + 1 M + 2 In other words, ||« is the integer nearest to |« -^ a DERANGEMENTS. 109 PROPOSITION XXXI. The number of ways in which a row of n elements may he so deranged that no element shall he in its proper place is \\ n, or the integer nearest to \n -=- e. For the total number of arrangements is In. The num- ber fulfilling the condition that any assigned one (at least) is in situ is In — 1. The number fulfilling the condition that two assigned elements are in situ is In — 2, and so on. Hence by Prop. XIV. Cor. 3, the number of arrangements in which none of the n elements are in situ is "in 1 1 '^('^-l) ■|n-2- .'^(-l)(-2)|,_3 + &,. ir 1' 1.2 to n + 1 terms 1 f-. 1 1 1 -j3+- ••±^} = M: Q-E-D. CoKOLLART. The number of ways in which n elements can be deranged so that not any one of r assigned elements ■may he in its proper place [the rest being unrestricted) is , r . ^ r(r — V) , „ , \n-j \n-l + ''^^ | n-2 -... + |n-n Example. Suppose we have the four elements a, h, c, d; the number of derangements, so that all may be displaced, is (by the proposition) [[4 of 9. : " " These nine derangements are as follows : b d a c c a db d c ah bade c db a d ch a b c d a c d ah d ah c 110 CHOICE.' If it be required to derange the same terms so that two may remain in situ and two be displaced, the number of derangements is, by the Corollary, 12(1 -1+1) = 6. These six derangements are as follows : ah d c a d cb a cb d b a c d chad db c a PROPOSITION XXXII. The number of ways of deranging a series of n terms so that no term may be followed by the term, which originally followed it, is \\n + \ \n — 1, or the integer nearest to \n + l -¥■ ne. Let a, j8, 7,...^ represent the n terms. Then amongst the \n arrangements of which the terms are capable there will be Ire — 1 in which any assigned sequence a/3 occurs : (for these arrangements will be obtained by regarding a/8 as one term and then arranging it with the remaining « — 2 terms). Similarly any two sequences which can consistently occur (as a/3, /S7, or a^S, yS*) will be found in |w — 2 different arrangements. Any three consistent sequences will be found in Iw. — 3 different arrangements : and so on. Therefore the number of arrangements free from any of the n — 1 sequences is, by Prop. XIV. Cor. 3, \n-{n-l) \n-l + ^ ^^ -\n-_2 — -^^ 'j- 'o^ ' \n — 3+ &c. to n terms * Of. course such sequences as a/3 07 could not consistently occur, as a could not at the same time he followed by j3 and 7. DERANGEMENTS. Ill I ^f n-1 n-2 ra-3w-4 „ 1) = l!izif- Y' H 1"''T~ y' which is easily seen to be equivalent , to jlw + ||?z — 1, or \\n + 1 -=- 71 (page 108), which is the integer nearest to |to+ 1 4- we. Q. E. D. Corollary. The nwmber of derangements of a series of n terms, free from any of r assigned sequences {capable of occurring simultaneously in one arrangement), is \n-^ \n-l + '^^^~^^ \n-2 -... + \n -r. Example. Let us derange the series of four elements abed so as to exclude the sequences ah be cd. By the proposition the number of derangements is [14 + p = 9 + 2 = ll. And they are found on trial to be a c b d b d c a c a db d b a c a d c b bade c b a d d c b a b d a c cb d a d a c b PROPOSITION XXXIII. ' The number of ways of deranging the series of n terms a, jS, 7, ...I, K, so that none of the n sequences a/S, /Sy, ...ik, Ka may occur, is n [|w— 1, or n times the integer nearest to \n-\^e.* * Or 1 more (n odd) or 1 lesa (n even) than the greatest integer in |«-=- e. 112 CHOICE. For, as before, there are \n arrangements of the n terms, and any assigned sequence will lie found in \n-'l of them ; any two assigned sequences in Ira— 2, any three assigned sequences in Iw — 3 of them, and so on for n—1 sequences ; but as all the n sequences cannot occur at .once, the number of arrangements with n sequences is not 1 but 0. Therefore (by Prop. XIV. Cor. 3) the whole number of admissible arrangements is , , _ ?i(m— 1), no |n-}i|K- 1 +— ^:j — ^ [w-2 -&c. to n terms [hot n+1 terms], which may be written n \\n— 1, which is n times the greatest integer in |w— 1 -i-e. q.e.d. Example. Let us derange the series of four elements abed so as to exclude the sequences ab bo cd da. By the proposition the number of derangements is 4 [jS or 4 X 2 = 8 ; and the eight derangements are found to be a cb d b d c a c a db db a c a d cb bade cb a d d cb a PROPOSITION XXXIV. Ifn terms be arranged in circular procession the number of ways in which they can be deranged so that no term may be followed by the term which originally followed it, is I fl_J_, _^^ 1 . ,11 \-\n .n-l^\2(n-2) |3(w-3)"^ •• - In] " DERANGEMENTS. 113 For the whole number of arrangements of n things in circular procession is |w — 1 ; and any assigned sequence is found in \n— 2 of them ; any two assigned consistent se- quences in Ift— 3,'and so on, till we come to the case of n consistent sequences, which can evidently be found in 1 arrangement. Hence provided we replace the final I— 1 by unity we may write down, (by Prop. XIV. Cor. 3), the number of arrangements free from any of the n sequences as follows ; In — 1 —7i\n — 2 -\ — ^ — ^ \n —3 — &c. to (n + 1) terms I I X , ^ ' _, fl 1 1 1 11 ~l-|w |1 (n - 1) "^ |2 (n - 2) \S(n-3y"-^]- Q.E.D. Examples : If n = 8, the number of derangements may be written |2-3|1 + 3|0-1 = 1, or the only available derangement is one, viz. the one in which the order of the terms is reversed. If «, = 5 we have |4 - 5 |3 + 10 |2 - 10 |1 + 5 |0 - 1 = 8. If abcde represent the original order, the eight derange- ments may be exhibited as foUows : ached a eb d c a c e db a e cb d a d c eb adbec acebd aedcb If re = 6 the number of derangements is 36 : if m = 7 it is 229: and if w = ll it is 1214673. And always if n be a prime number, the number of derangements, with 2 added, is a mult^le of n. w. 8 114 CHOICE. In the foregoing propositions we have investigated the number of ways of deranging groups of elements subject to various laws. But as there can scarcely be a limit to the variety of laws which might be proposed to regulate the dis- tribution in different cases, it would be an endless task to undertake a strictly complete discussion of the subject, or to make our treatise exhaustive. The cases which we have considered are those which most obviously arise, and the methods which we have applied to them will be easily adapted to a variety of other cases, or will suggest other methods of still wider applicability. To save trouble in working out numerical examples we append a table of the values of the first 12 factorials and sub-factorials, and of their ratios. It will be seen that the ratio of l« : [|ri converges with great rapidity towards its ultimate value e. n. l» ]\n |«-^|jm [[».-=-(» 1 1 00 0-000000 2 2 1 2-000000 0-500000 3 6 2 3-000000 0-383833 4 24 9 2-666666 0-375000 5 120 44 2-727272 0-366666 6 720 265 2-716981 0-368055 7 5040 1854 2-718446 0-367857 8 40320 14833 2-718262 0-367881 9 362880 133496 2-718283 0-367879 10 3628800 1334961 2-718281 0-367879 11 39916800 14684570 2-718281 0-367879 12 479001600 176214841 2-718281 0-367879 EEEATA. Page 108 lines 3, 4 &om bottom odd, , even, for read ,, -^ even, odd. Page 111 footnote for greatest integer in read integer nearest to. CHAPTER V. SOME QUESTIONS OP PEIOKITT. We shall consider in the present chapter some questions of order amongst things of two or more classes, when limits are placed upon the excess at any point of things of one class above things of the other. If a man possessed of only £5 makes 24 successive bets of £1, winning 12 times and losing 12 times, the number of orders in which his successes and failures can occur will naturally be limited by the fact that the losses must never exceed the gains by more than £5. A variety of questions may be proposed in which similar conditions are assigned. For example, if an urn contain black balls and white balls which are to be drawn out in succession, the order in which they are drawn out may be limited by the condition that the number of white balls drawn must never exceed the number of black balls ; or, again, by the condition that the excess of black balls over white must never be more than a given number. Or there may be a question of taking m paces horizon- tally and n paces vertically without ever crossing that oblique line which is reached when the number of paces in each direction is the same. This last case suggests a graphic representation which will equally apply to the other cases, and will help to fix our 8—2 116 CHOICE. ideas. The horizontal paces may be taken to represent successes and the vertical paces failures ; or the one may represent the drawing of black balls and the other of white balls. In stating our propositions we shall speak for short- ness of gains and losses, but it will be observed that the reasoning is quite general. From any origin draw OX horizontally and F vertically, and let P be the point which would be reached by starting from and taking m paces horizontally in the direction OX and n paces vertically parallel to OT. Let A be the point reached by taking one pace hori- zontally and one pace vertically, B the point reached by taking two paces in each direction, C by taking three paces in each direction, and so on; so that ABG... lie in the straight line bisecting the right angle XOY. Let 0, a, h, c... be the points reached by taking one horizontal pace from each of the points 0, A, B, G... and. ca, a, /8, 7. . . the points reached by taking one pace vertically from 0, A, B, C... respectively. QUESTIONS OF PEIOKITY. 117 la considering the number of routes from one point to another we shall proceed on the understanding that each route is to be traversed by paces taken only horizontally and vertically, and without retrogression. It is plain that the whole number of routes from to P is Cy or C„ * ' m, n' PEOPOSITION XXXV. G„„ being the number of orders in which n gains and n losses can occur, and J^„ denoting the number of orders which fulfil the condition that the losses must never be in excess of the gains, C'>.. » + or, ^n,n + "A, 1 ^«-a, n-1 + «^2, 2 C^n-2 n-2 + • • • ■ • • + "^n-l. n-1 C'l, 1 = 2 {G„ „ — J^ „). Q.E.D. PROPOSITION XXXVI. Of the orders in which n gains and n losses can occur, ths of the whole number will fail to fulfil the condition n + 1 •' that the losses must never exceed the gains. The number of orders in which n gains and n losses can occur is (7„ „. ,With the notation already adopted we have to prove that n _ r — ^ c or that "%i~^> "3,3-— ^> "4,4~^^- Hence the proposition is true as long as n does not exceed 4. To prove it generally, we shall shew that if it be true when n = 0, 1, 2, ... or a; — 1 it wUl also be true when n = x. By the last proposition we have = 2(7^„ — 2,/^„. But since our present proposition is by hypothesis true when J? = 0, 1, 2 ... or a; — 1 we have T — f " 0, ^0, 0' "2,2 ~ 3W21 &C. = &C. 1 J.. G,. Substituting these values we get + ~ ^i-l, ar-l C'l, 1 — 2(7^^ , — J^ or (Excursus, Art. 13, page 82) Gx j;,, — «:. ^1 ^ + 1 Cv , — 2(L_ _ — 2 JL Therefore 2/..= 2 + ^^ 2a! + r a; + l x+ 1 C. /.,= 120 CHOICE. that is, if the proposition is true for the first x—l values of n, it is true for the next value. But we have found that it was true for initial values ; therefore universally "■ " ^ m + 1 "•"' and C^„ „ — Jl „ = ^ G^ Note. The result may also be written T = C — O "n,n ^n,n ^n+1, m-1> and G^„— J^„= C„^^_ „.i. LEMMA. • To sum the series Go. (,^m,«+ h^l, 1 ^m-1, »-l + i^2, 2 C'm-a, «-2 + ■ ■ Q. E. D. I -*■ ri ft In the diagram already constructed, since all the routes from to P commence at the point on the diagonal and terminate at the point P to the right of the diagonal, they may be classified according to the points at which they first pass to the right of the diagonal : i.e. according as they first pass along Oo, or ^a or Bb, &c. Consider those which pass along Kk. Any such route consists of three portions OK, Kk, kP, of which OK lies altogether to the left of the diagonal, and can therefore be described in r-Cy, ways (Prop. XXXVI.); Kk can be described in one way, and kP can be described in C„_,.i „., ways. Hence there are QUESTIONS OF PEIOEITY. 121 such routes ; and giving r all values from.O to n inclusive we must obtain all the C„ „ routes from to P. Therefore we • must have identically This is true for all values of m and n ; therefore, writing '»i + 1 for TO we obtain Note. If we make m — n, we obtain a series already sumihed (Excursus, Art. 13, page 82). PROPOSITION XXXVII. ^m. n being the number of orders in which m gains and n losses can occur, and Jm.n denoting the number of orders which fulfil the condition that the losses must never he in excess of the gains, (^m, n-l + "A, fitn-\, n-2 + '^2, 2^ra-2, >.-3 + ■ • • I 7" n — r — T • • ■ "r ^ n-l, n-r^m-TH-X ^m, n ^ m, n- , Proceeding step by step as in Prop. XXXV, but con- sidering the routes from to P instead of to F we find that each route OKkP may be made in J^.fi,^ ■'m.-r, n-r-1 ways. Hence ^r=0 ly r, r m-r, n-r-li ^m, n m, n or ^m. n-l + ''l, 1%»-1, n-2 + ^1, 2^ra-2, m-3 + • • • • • • + " 11- 1. n-r-'m-n+l, ~ ^m, n " m, n- Q. E. D. 122 CHOICE. PROPOSITION XXXVIII. Of ihs orders in which m gains a/nd n losses can occur — —^ ihs of the whole number will fail to fulfil the condition that the losses mv,st never exceed the gains. We found in Prop. XXXVI. the values of Jj ,,, J^i, J^,^, &c. in the form J -- ^ C "• " re + 1 "■"■ Substitute these values in the result of Prop. XXXVII. and we obtain ^m, n-l + 2 ^1, l^OT-1, n-2 ■"■ 'S ^2, 2^m-2, n-S + • • • '" m "-1. l-l ^m-m+1, ^^ ^m, » " m. n- But by the Lemma (page 120) writing n — 1 for m in its result, the first member of this equation is equal to (7,^^ „.i. Therefore n _ r _ /-» ^m, n "m, ■» '-'nl+l, n-l m + 1 "*■ "■ Q. E. D. ^ r m — n + 1' Corollary. J;„, „ = ^_^^ C„, „. EXTENSION OF THE LEMMA. If in the proof of the Lemma (page 120) the routes instead of starting from had started from the point F situated h paces horizontally to the left of 0, the whole number of routes would have been G„+„^„. And classifying^ them as before, any route passing along Kk would have consisted of three parts, FK, Kk, kP, of which FK could be QUESTIONS OF PRIORITY. 123^ described in ^ ^^^ ^ C^,,, ways (Prop. XXXVIII.), Kk ia one way, kP in (7„.,.i, „., ways. Hence there are ^q-^-j-y C,„.,.i „., C,+,,, such routes. And giving r all values from to w we obtain the identity ^^2 *+'•"'"-='■ "-^"^ A, + 3 , "'+-'■ fi p _ p ^t, 0> ^>»-l> n "■" I, I n ^M-1, 1' ^™-2, n-1 "■" I I Q ^M-2, 2' ^m-3, n-S + ■ • • or h + 1 "^ /n-2 "^ h + S "^•■" PC o ■■■ A + M+l A+1 "^ Write A for A + 1 and m for m - 1, and we have h h h + \ '" h + n Now write m — A, for n, and we have p f< p p p P P A A A+ 1 "■ w ' PROPOSITION XXXIX. Among the (7„ „ orders in which m gains and n losses can occur, there will he G.„^^„_^ orders in which the losses will at some time be at least h in excess of the gains. With the same diagram as before, the orders wiU be ■ represented by routes from to P which touch or cross a line HR parallel to the diagonal OK, at a vertical distance of h paces above it. The routes may be classified according to the points at which they first touch the line HR. 124 CHOICE. Let Ox'LP be any route which first reaches this line at a point L, distant r horizontal paces and r + A vertical paces from 0, and let "k'L be. the pace by which the point L is approached. The route may be divided into three parts OX', X'i, and LP; of which h Ox' can be made in ^ (^r+h-i.r'^^J^ (Prop. XXXVIII.), \'L in one way, LP in (7„_,, „.,.^ ways ; therefore the number of routes first touching at L is and the whole number of routes will be got by giving r all values from to n — h inclusive, and addiDg. The summation is that of the final series of the extension of the Lemma (page 123). Hence, the whole number of routes is C',^^ „_j, which therefore represents the number of orders required. Q. E. D. CoEOLLAEY. It follows that the number of orders in which m gains and n losses can occur so that there shall never be an excess of h gains, is G^^ „ — G,^^ „_,,. Similarly, the number of orders so that there shall never be an excess of k losses, is G^^ „ — G^_^^ ^^. If A = 1 this reduces to the case of Prop. XXXVIII. CHANCE. CHAPTER VI. THE THEORY OF PROBABILITY TREATED ARITHMETICALLY. " There is very little chance of fine weather." " Is there much chance of his recovery ? " " There is no chance of finding it." " There is a great probability of war." " This is a more probable result than the other." " That is more likely to be mine than yours." " There is less chance of her coming than of his.'' These are expressions in common use amongst us; the very commonness of their use shews that people in general have some idea of chance, and some conception of different degrees of probability in the occurrence of doubtful events. All understand what is meant by much chance and little chance ; they distinguish events as very probable, probable, improbable, or very improbable ; but no attempt is made in common conversation to measure with any accuracy the amount of probability attaching to any given event. If a Doctor is asked what chance there is of a patient's recovery, he may answer that there is much chance or little chance, but he cannot express with any precision the exact magni- tude of his hope or of his fear. Yet his expectation of the 126 CHANCE. event has a certain magnitude. He has a greater expecta- tion of this patient's recovery than he has of the recovery of another, whose symptoms are more aggravated, and less expectation than in another case where the constitution is stronger. His expectation has a definite value, and if he were a gambler, he would be prepared to offer or take certain definite odds on the event. But in common lan- guage, this definite amount of expectation or probability cannot be precisely expressed, because we have no recognised standard with which to compare it, no recognised amount of expectation or probability by which to measure it. In fact, in describing the magnitude of any expectation which we entertain, we are in the same position as if we had to describe the length of a room, or the height of a tower, to a man who was not acquainted with a foot or a yard, or any of our standards of length. We could speak of the room as very long or very short, we could speak of the tower as very high or very low, but without some standard length recognised alike by ourselves and those whom we addressed, we could not give an accurate answer to either of the questions. How long is the room ? or How high is the tower ? So when we are asked what chance we think there is of a fine afternoon, we may say that there is much chance or little chance, or we may even go further, and establish in our own minds a scale of expressions, distinguishing the dif- ferent degrees of probability in some such way as follows : It is certain not to rain. It is very unlikely to rain. It is unlikely to rain. It is as likely to rain as not. It is likely to rain. It is very likely to rain. It is certain to rain. CHANCE IMPLIES IGNORANCE. 127 But these expressions, except the first, fourth, and last, are vague and indefinite, nor can we ever be sure that those ■with whom we are conversing attach exactly the same idea to each expression that we do. This vagueness is of little consequence in common Hfe, because in most cases it is impossible to make an accurate estimate of a chance, and the expressions are, perhaps, as accurate as the estimates themselves which we wish to express. But there are other classes of events concerning which it is possible to form accurate estimates of their degree of prbbability or likelihood of happening, and in these cases it is well to have some more precise method of expressing diiferent degrees of probability, than is afforded hy the common expressions which we have quoted. We must observe at the outset, that we use the words chance and probability as strictly synonymous. In common language, it is usual to prefer the former word when the •expectation is small, and the latter when it is large. Thus we generally hear of "little chance," or of "great proba- bility," but not so often of " great chance," or " little proba- bility." This distinction, however, is not universal, and we shall entirely disregard it, using the two words chance and probability in the same sense. It will be seen that probability always implies some ignorance on the part of the person entertaining the expec- tation, and the amount of probability attaching to any event will depend upon the degree of this ignorance. With omniscience, degrees of probability are incompatible; for omniscience implies certainty, and certainty precludes doubt, and degrees of probability are the measures of doubt. Hence, there is no such thing as the absolute probability of an event, all probability being conditional on our igno- rance, and varying when that condition varies. Thus the 128 CHANCE. same event will be unequally probable to different persons, whose knowledge of the circumstances relating to the event is different. And to the same person, the expectation of any event will be affected by any accession of knowledge concerning the event. For instance, suppose we see a friend set out with five other passengers in a ship whose crew number thirty men : and suppose we presently hear that a man fell overboard on the passage and was lost. So long as our knowledge is con- fined to the fact that one individual only has been lost out of the thirty-six on board, the probability that it is our friend is very small. The odds against it would be said to be thirty-five to one. But suppose our knowledge is augmented by the news that the man who has been lost is a passenger ; though we still feel that it is equally likely- to be any of the other five passengers, yet our apprehension that it is our friend becomes much greater than it was before. The odds against it are now described as five to one. Thus the proba- bility that our friend is lost is seen to be entirely conditional on the respective degrees of our knowledge and ignoraijce ; and so soon as our ignorance vanishes — so soon as we know all about the event, and become as far as that event is con- cerned omniscient, — then there no longer remains a question of probability; the probabiUty is replaced by certainty. This example will also illustrate the meaning of the rsttio of probabilities. Since each of the passengers was equally likely to have been lost, it was evidently always six times as likely that the man lost was some passenger, as that it was our friend. So it was five times as likely that it was a passenger, but not our friend, as that it was our friend. Therefore, also, the probability that it was a passenger, but not our friend, was to the probability that it was a passenger in the ratio of 5 to 6. RATIO OF PROBABILITIES. 129 Let US suppose another case. A number of articles are placed in a bag, and amongst them are three balls, aUke in all respects, except that two of them are coloured white and the third black : all the other articles we will suppose to be coins, or anything distinguishable without difficulty from We present this bag to a stranger, and we give him leave to put in his hand in the dark, and to take out any one article he Ukes. But before he does this, we may consider what chance there is of his taking out a ball, or what chance there is of his taking out the black ball. Obviously we cannot form any accurate estimate of this chance, because it must depend upon the wants or the taste of the stranger influencing his will, whether he will prefer to take a ball or a 'coin, and being ignorant of his will in the matter, we cannot say whether it is likely or unlikely that he will select a ball. But it is axiomatic, that if he draws a ball at all, it is twice as likely to be a white ball as to be a black one, or the respective chances of his drawing white or black are in the ratio of 2 to 1, and these chances are respectively two- thirds and one-third of the' chance that he draws a ball at all. We now proceed to shew how the magnitude of a chance ml^ be definitely expressed. We have already pointed out that the expressions used in common language are wanting in definiteness and precision, and we compared the expe- dients by which degrees of probability are usually indicated to the attempts which we should make to give an idea of the length of a room to a person unacquainted with the measures of a foot and a yard. Now we observe, that the difficulty in this latter case ceases, so soon as the person with whom we are speaking w. 9 130 CHANCE. agrees with us in his conception of any definite length what- ever. If he can once recognise what we mean by the length of a hand, for instance, we can express to him with perfect accuracy the length of the room as so many hands ; or, if he have an idea of what a mile is, we can precisely express the length of the room as some certain fraction of a mile. So, also, as soon as we have fixed upon any standard amount of probability that can be recognised and appreciated by all with whom we have to do, we shall be able to express any other amount of probability numerically by reference to that standard. The numbers 2, 3 would express probabilities twice or three times as great as the standard probability; and the fi-actions i,^, | would express probabilities half, one-third, or two-thirds of the standard. Now, it matters not how great or how small the standard be, provided it be a probability which all can recognise, and which all will alike appreciate. This is, indeed, the one essential which it has to fulfil; it must be such that all persons will make the same estimate of it. And that which best satisfies this condition, and, therefore, the most con- venient standard with which to compare other probabilities, is that supreme amount of probability which attaches to an event which we know to be certain to happen. All under- stand what certainty is : it is a standard which all estimate alike, Certainty, therefore, shall be our unit of probability; and other degrees of probability shall be expressed as frac- tions of certainty, But it may be asked. Is certainty a degree of probability at all, or can smaller degrees of probability be said to have any ratio to certainty? Yes. For if we refer to the instance already cited of the six passengers in the ship, we observe that the chance of the lost man being a passenger is six times as great as the chance of his being our friend. This CERTAINTY A MEASXTEE OF PROBABILITY. 131 is the case however great our ignorance of the circumstances of the event; and it ■wUl evidently remain true until we attain to some knowledge which affects our friend differently from his fellow-passengers. But the news that the lost man was a passenger does not affect one passenger more than another. Therefore, after receiving this news, it will still hold good that the chance of the lost man being a passenger is six times as great as the chance of its being our friend. But it is now certain that the lost man was a passenger; therefore the probability that it was our friend is one-sixth of certainty. Again, in the instance of the balls and coins in the bag, we have already noticed that the chances of drawing white or black are respectively two-thirds and one- third of the chance of drawing a ball at all. And this is the case whatever this last chance may be. But suppose the man tells us that he is drawing a ball, not a coin, then this last chance becomes certainty; and therefore the chances of drawing white or black, become respectively two-thirds and one-third of certainty. Thus it is seen that certainty, while it is the supreme degree, is some degree of probability, or is such that another degree of probability can be com- pared to it and expressed as a fraction of it. Of course, when we use unity to express certainty, the probability of the lost passenger being our friend wUl be expressed by the fraction g, and the chances of the ball drawn being white or black, will be expre.ssed by the frac- tions 3 and 3. After the explanations which we have already given, the reader will have no difficulty in accepting the following 9—2 132 CHANCE. AXIOM. If an event can happen in a number of different ways (of which only one can occur), the probability of its happening at all is the sum of the several probabilities of its happening in the several ways. For instance, let the event be the falling of a coin. It can fall either head or tail, and only one of these ways can occur. The probability that it falls at all must be made up by addition of the probability that it falls head and the probability that it falls tail. Again, let the event be that either A, B, or G should win a race in which there are any number of competitors. The event can happen in three ways, viz., by A winning, by B winning, or by G winning ; and only one of these ways can occur. The probability that one of the three should win is equal to the sum of the probabilities that A should win, that B should win, and that G should win. This is only saying that if a man would give £2 for A's chance of the prize, £3 for B's chance, and £4 for G's chance, he would give £2 + £3 + £4, or £9 for the promise that he should have the prize if any one of the three should win. Again, if jq be the chance of a shot aimed at a target hitting the bull's eye, g the chance of its hitting the first ring, and j the chance of its hitting the outer ring, the chance that it hits one of these, i.e., the chance of its hitting the target at all, is jq + g + j , or gg . EULE I. The probability of an event not happening is obtained by subtracting from imity the probability that it will happen. COMPLEMENTARY PROBABILITIES. 133 For it is certain that it will either happen or not happen, or the probability that it will either happen or not happen is unity ; and only one of these two (the happening and the not happening) can occur. Therefore, by the axiom, unity is the sum of the probabilities of the event happening and not happening; or the probability of its not happening is obtained by subtracting from unity the probability of its happening. - 2 Question 83. If the chance of an event happening is g , what is the chance of its not happening ? Answer. 1 — 5 > or g . Question 84. If the chance of an experiment succeeding 7 3 IS j^ , what is the chance of its failing ? Answer, jg . Question 85. If the chance of a shot hitting a target 31 29 be gg , what is the chance of its missing ? Answer, gg . Question 86. If the chance of A winning a certain race be g, and the chance of B winning it 5, what is the chance that neither should win ? 17 Answer, ht . For, by the axiom, the chance that one of 117 them should win is § + g , or gj ; and therefore, by Rule I., 7 17 the chance that this should not happen is 1 — 57 , or 57 . Definition I. Two probabilities which together make up unity, are called complementary probabilities. Definition II. When it is said that the odds are three to two against an event, it is meant that the chance of the event failing is to the chance of its happening as three to 134 CHANCE. two ; and when it is said that the odds are three to two in favour of an event, it is meant that the chance of its happen- ing is to the chance of its failing as three to two ; and so for any other numbers. RULE II. If the odds be three to two against cm event, the chance of the event not happening is 3 3 + 2' and the chance of its happening is 2 3 + 2' a/nd so for any other nwmbers; the nii/merators of the two fractions being the two given numbers, and their common denominator the swm of the nwmbers. For the two fractions satisfy the condition required by Rule I., viz., that their sum should be unity, and that required by the definition, viz., that their ratio should be the same as the ratio expressing the given odds. Similarly, If the odds be three to two in favour of an event, the chance of the event happening is 3 3 + 2' and the chance of its not happening is 2 3 + 2' and so for any other mimbers; the numerators of the two fractions being the two given numbers, and their common denominator the swm of the nwmbers. FUNDAMENTAL EULE. 135 Question 87. If the odds be ten to one against an event, what is the probability of its happening ? Answer, j^. Question 88. If the odds be five to two in favour of the success of an experiment, what are the respective chances of success and failure ? Answer. The chance of success is s and the chance of failure ^ . RULE III. If am, event can happen in five ways, and fail in seven ways, and if these twelve ways are all equally probable, and only one of them can occur, the odds against the event are seven to five, and the chances of its happening and failing are respectively 12 '""^12' a/nd similarly for any other numbers. For since the event must either happen or fail, one of the twelve ways must occur;, therefore the sum of their several probabilities is unity. But all the twelve ways are equally probable. Therefore the chance of the occurrence of any particular one is jg > and the chance of the occurrence of one of the five which cause the event to happen is five times this, or j2 • ^° *^® chance of the occurrence of one of the 7 seven which cause the event to fail is ^ . Suppose, for example, that a die has twelve faces, of which five are coloured white and seven black. A person throws the die, and is to receive a prize if it faU white. 136 CHANCE. The odds are seven to five against his winning the prize. 12' 5 The chance that he wins is j^ , and the chance that he loses ■ 7 lSj2. For all the twelve faces are equally likely to turn up, and one must turn up. Therefore the chance of any particular face turning up is j2 > and the chance of a white face turning 5 up is five times this, or j2 • Or we might put it thus: Since there are five white and seven black faces, it is axiomatic that the chance of a white face is to the chance of ai black face as five to seven. Now ds soon as it is certain that the die is to be thrown, it is certain that either a white or a black face must turn up. The two chances must therefore now make up unity. But they still retain the ratio of five to seven, therefore they become respectively and 5 + 75 + 7' And in the same way we might reason if the numbers were any other. Question 89. A party of twenty-three persons take their seats at a round table ; shew that the odds are ten to one against two specified persons sitting together. Answer. Call the two specified persons A and B. Then besides A's place (wherever it may be) there are twenty-two places, of which two are adjacent to A's place and the other twenty not adjacent: And B is equally likely to be in any of these twenty-two places. Therefore (Rule III.) the odds are twenty to two, or ten to one, against his taking a place next to A. FUNDAMENTAL BULK. 137 The last rule may be expressed in a somewhat different form as follows : RULE IV. If there he a number of events of which one must happen and all are equally likely, and if any one of a {smaller) number of these events will produce a certain result which cannot otherwise happen, the probability of this result is expressed by the ratio of this smaller number to the whole number of events. For instance ; if a man has purchased five tickets in a lottery, in which there are twelve tickets altogether and only one prize, his chance of the prize would be expressed by the ratio 5 : 12, or by the fraction ^. For convenience of reference we have given distinct num- bers to the two Rules III. and IV., although they are only different statements of one and the same principle. This will be immediately seen, by considering the case of the lottery just instanced. We might at once have said that there were twelve ways of drawing a ticket, and five of these would cause the man to win, while the other seven would cause him to lose. Rule III. is therefore immediately ap- plicable. Question 90. If from a set of dominoes numbered from double-blank to double-twelve one domino be drawn, what is the chance that one of the numbers on it is 12 ? 13 14 Answer. There are ' , or 91 dominoes in the set X . ^ (Choice, Rule X.), and the number twelve appears on 13 of them, since it will appear in combination with blank as well as with each of the twelve numbers. Hence there 138 CHANCE. are 91 equally probable events of which 13 will produce the result that a twelve is drawn. Therefore the chance is 13 1 910^7- Question 91. The four letters s, e, n, t are placed in a row at random ; what is the chance of their standing in such order as to form an English word 1 Answer. The four letters can stand in |4 or twenty-four different orders (Choice, Rule III.): all are equally likely and one must occur. And four of these will produce an English word, sentj nest, nets, tens. 4 1 Hence by the rule, the required chance is 24 or g . Question 92. What is the chance of a year, which is not leap year, having fifty-three Sundays ? Answer. Such a year consists of fifty-two complete weeks, and one day over. This odd day may be any of the seven days of the week, and there is nothing to render one more likely than another. Only one of them will produce the result that the year should have fifty-three Sundays. Hence (Rule IV.) the chance of the year having fifty-three Sundays is ^ • Question 93. What is the chance that a leap year, se- lected at random, will contain fifty-three Sundays ? Answer. Such a year consists of fifty-two complete weeks, and two days over. These two days may be Sunday and Monday, Monday and Tuesday, Tuesday and Wednesday, Wednesday and Thursday, EXAMPLES OF RULES IIL AND IV. 139 Thursday and Friday, Friday and Saturday, Saturday and Sunday, and all these seven are equally likely. Two of them (the first and last) will produce the required result. Hence 2 (Eule IV.) the chance is ^ . Question 94. What is the chance that a year which is known not to be the last year in a century should be leap year ? Answer. The year may be any of the remaining ninety- nine of any century, and all these are equally likely; but twenty-four of them are leap years. Therefore (Eule III.) the chance that the year in question is a leap year is gg or gg. Question 95. Three baUs are to be drawn from an urn which contains five black, three red, and two white balls. What is the chance of drawing two black balls and one red ? Answer. Since there are ten balls altogether, three balls can be drawn in ' " , or 120 different ways, all equally 5 4 likely. Now, two black balls can be selected in =-^ , or ten ways, and one red in three ways. Hence, two black balls and one red can be drawn in 10 x 3, or 30 different ways. Thus we have 120 different ways of drawing three balls, whereof 30 ways will give two black and one red. Hence, when three balls are drawn the chance that they should be two black and one red is (by Rule IV.) ! 30 1 120 ""4- 140 CHANCE. Question 96. If from a lot of thirty tickets, marked 1, 2, 3, &c., four tickets be drawn, what is the chance that those marked 1 and 2 are among them ? Answer. Four tickets can be drawn out of thiiiy in 30.29.28.27 i? r i . i. a — 5 — ^ ^ , — ways, iour tickets can be drawn, So as to 1.2.3.4 •' 28 27 include those marked 1 and 2, in -y—o ways. Hence, when four are drawn, the chance that these two are included is 28 ■ 27 . 30 ■ 29 . 28 . 27 _ 3.4 _ 2 1.2 ■ 1.2.3.4 "29.30 "145" The odds are, therefore, 143 to 2 against the event. Question 97. A has three shares in a lottery where there are three prizes and six blanks. B has one share in another, where there is but one prize and two blanks. Shew that A has a better chance of winning a prize than B, in the ratio of 16 to 7. Answer. A will get a prize unless his three tickets all 9 8 7 prove blank. Now, three tickets can be selected in ' ' , or 84 ways; and they can be selected so as to be all blank 6.5.4 in o q » °^ 20 ways. Hence the chance that they should 20 5 be all blank is ttt or ^jv ; and, therefore, the chance that this 84 21 5 should not be so, or thaft A gets at least one prize, is 1 — ^, or n^ • But it is evident that the chance that B gets a prize is 1 7 (Eule IV.) Q or jr:r . Therefore A has a better chance than B in the ratio of 16 to 7. EXAMPLES. 141 Question 98. If four cards be drawn from a pack, what is the chance that there will be, one of each suit ? Answer. Four cards can be selected from the pack in ^^i'^l'l^A^^ OT 270725 ways (Choice, Eule IX.); but four cards can be selected so as to be one of each suit in only 13 X 13 X 13 X 13 or 28561 ways (Ghoice, Rule II.). Hence the chance is 28561 ,.,^, ,, 1 or a little more than 270725 10 ■ Question 99. If four cards be drawn from a pack, what is the chance that they will be marked one, two, three, four ? Answer. There are 4 x 4< x 4 x 4, or 256 ways of draw- ing four cards thus marked, and 270725 ways of drawing four cards altogether. Hence the chance is 256 270725 ' or the odds are more than 1000 to 1 against it. 100. In a bag there are five white and four black balls. If they are drawn out one by one, what is the chance that the first will be white, the second black, and so on alternately ? Answer. The nine balls can be arranged in 19 ways. But the five white balls can be arranged in the odd places, and the four black balls in the even places in |5 x 14 ways. Hence the chance that the order will be the alternate order is|5x^-^^orjL. Otherwise. The white balls being for the purposes of the question all alike, and the black balls all alike, the total 142 CHANCE. number of orders in which the nine balls can be arranged is (Choice, Eule VII.) -!^, or 126. |4.|5 The balls are equally likely to be drawn in any of these orders, and one of them is the alternate order required. Therefore the chance is ^sr; • 126 Question 101. In a bag are five red balls, seven white balls, four green balls, and three black balls. If they be drawn one by one, what is the chance that all the red balls should be drawn first, then all the white ones, then all the green ones, and then all the black ones ? Answer. The nineteen balls can be arranged in 119 [5 . 1 7 . 1 4 . 1 3 different orders (Choice, Rule VII.). AU these are equally likely, and therefore the chance of any. particular order is 1 5 . (7 . 1 4 . |3 ^9 ' This will be the chance required ;. for, all individuality among balls of the same colour having been disregarded, only one of the different arrangements will give the order of coloiirs prescribed in the question. Question 102. Out of a bag containing 12 balls, 5 are drawn and replaced, and afterwards 6 are drawn. Find the, chance that exactly 3 balls were common to the two drawings. Answer. The second drawing could be made altogether in 112 -^=^, or 924 THROWS WITH TWO COMMON DICE. 143 ways. But it could be made so as to include exactly 3 of the balls contained in the first drawing, in |5 |7 '- x-!^-, or 350 [3.|2 [3.|4 ways ; for it must consist of a selection of 3 balls out of the first 5, and a selection of 3 balls out of the remaining 7 (Choice, Rules VIII. and II.). Hence, the chance that the second drawing should contain exactly 3 balls common to the first, is 92J or gg . As the respective probabilities of various throws, with two common dice, are of practical interest, in their bearing upon such games as Backgammon, it may be well to discuss this case with some completeness. It will be observed that as each die can fall in six ways, the whole number of ways in which the two dice can fall is 6 X 6 or 36. But these 36 different ways are not practically different throws, since, for example, it makes no difference in practice whether the first die falls six and the second five, or the first five and the second six. The number of practically different throws is, in fact, only =-^ or 21 ; being the num- her of ways of selecting two numbers out of six numbers when repetitions are allowed: {Choice, Rule X.). Regarding it in another way, the 36 different ways of the dice faUing are made up of sis unique ways — 1 and 1, 2 and 3, 3 and 3, 4 and 4, 5 and 5, 6 and 6, and 30 other ways, consisting of 15 essentially different throws, each repeated twice : thus — 1 and 2, 1 and 3, 1 and 4, 1 and 5, 1 and 6, 2 and 1, 3 and 1, 4 and 1, 5 and 1, 6 and 1, 144 CHANCE. 2 and 3, 2 and 4, 2 and 5, 2 and 6, 3 and 2, 4 and 2, 5 and 2, 6 and 2, 3 and 4, 3 and 5, 3 and 6, 4 and 3, 5 and 3, 6 and 3, 4 and 5, 4 and 6, 5 and 4, 6 and 4, 5 and 6, 6 and 5. Since each die is equally likely to fall in all different ways, the 36 different ways of the two dice falling are all equally likely; and, therefore, when the dice are thrown the proba- bility of any particular way is gg. But it cannot be said that all throws are equally probable, because six-five results practically in two ways out of the 36 ways of the dice falling, whereas six-six results in only one way. The correct state- ment is, that the probability of any assigned throw is gg if that assigned throw be dovhlets; but it is twice as much or jg if the assigned throw be not doublets. Thus the chance of throwing six-three is jg , but the diance of throwing three- three is gg . Question 103. When two dice are thrown, what is the chance that the throw will be greater than 8 ? Answer. Out of the 36 ways in which the dice can fall there are ten which give a result greater than 8, viz.: 3 and 6, ' 4 and 5, 4 and 6, 5 and 6, 5 and 5, 6 and 3, 5 and 4, 6 and 4, 6 and 5, 6 and 6. Hence the required chance is gg or jg . TWO COMMON DICE. 145 Question 104. What is the chance of throwing at least one ace ? Answer. Of the thirty-six ways in which the dice can fall, eleven give an ace. Hence the chance is gg . Question 105. What is the chance of making a throw which shall contain neither an ace nor a six ? Answer. Of the thirty-six ways, there are sixteen which involve neither one nor six. Hence the chance is gg or „ . This question, as well as the preceding one, may be more conveniently solved by Rule VI. Question 106. What are the odds against throwing doublets ? Answer. Of the thirty-six ways in which the dice can fall, six give doublets. Therefore the chance for doublets is gg or g , and the chance against doublets g (Rule III.). There- fore the odds are five to one against doublets. Or we might reason thus : — However the first die fall, the second die can fall in six ways, of which only one way will give the same number as on the first die. Hence, the odds are five to one against the second die falling the same way as the first, or the odds are five to one against doublets. Questimi 107. In one throw with a pair of dice, what is the chance that there is neither an ace nor doublets ? Answer. The dice can fall in thirty-six ways, but in order that there may be neither an ace nor doublets, the first die must fall in one of five ways (viz. 2, 3, 4, 5, 6), and the second, since it may be neither an ace nor the same as the &cst, may fall in four ways. Hence the number of ways w. 10 146 CHANCE. which will produce the required result is 5 x 4 or 20. And, 20 5 therefore, the chance of this result is gg or 5 . • Questmi 108. What is the chance of throwing exactly eleven ? Answer. Out of the thirty-six ways, there are two ways which produce eleven; therefore the chance is gg or jg . On the principle of the last answer, the reader will have no difl&culty in verifying the following statements : In a single throw with two dice, the odds are — 35 to 1 against throwing 2, 17 .to 1 n )> 3, 11 to 1 >j J) 4, 8 to 1 )» )i 5, 6|. to 1 >» » 6, 5 to.l » » 7, 6i to 1 J) 3) 8, 8 to 1 » „ 9, 11 to 1 J) )) 10, 17 to 1 )? )J 11, 35 to 1 11 1) 12. Thus the most frequent throw will be seven. In some cases the purpose of a throw is equally answered, whether an assigned number appear on one of the dice, or whether the numbers on the two dice together make it. Let us consider, for example, the chance of throwing five in this way. The chance of making a throw so that one die shall turn up five is gg , and the chance of making a throw which shall amount to five is gg. Therefore the chance of throwing five in one or these ways is gg + gg or gg . THREE COMMON DICE. 147 On this principle the following statements may be easily verified. In a single throw with two dice, when the player is at liberty to count either the smn of the nwmhers on the two dice, or the number on either die alone, the odds are — 25 to 11, or 2A to 1 against throwing 1, 24 to 12, or 2 to J ij 2, 23 to 13, or m to ) » 3, 22 to 14, or n to > J) 4, 21 to 15, or If to f jj 5, 20 to 16, or li to 7 it 6, 5 to J J) 7, 31 to 5, or H to J J) 8, 8 to ) J) 9, 11 to > }J 10, 17 to J >» 11, 35 to ) )J 12. Thus the number which there is the greatest chance of making is six. When three dice are thrown the number of ways in which they can fall is 6 x 6 x 6 = 216, but the number of 6 7 8 different throws is only ' ' or 56 {Choice, Rule X.). These 56 throws are made up of 6 triplets each occurring in one way, 30 throws containing a doublet, each throw occuning in 3 ways, 20 throws consisting of three different numbers, each throw occurring in 6 ways. And the chance of any given triplet (such as 6, 6, 6) is gjg- 10—2 148 CHANCE. 3 The chance of any given doublet (such as 6, 6, 5) is gjg 1 or^. And the chance of any other given throw (such as 6, 5, 4) • A. i ^® 216 °^ 36 • fi 1 Further, the chance of throwing some triplet is 2W^'^ 36' 90 • 5 The chance of throwing sOTue doublet is gie "'^ l2 • The chance of throwing neither a triplet nor a doublet is 120 5 216 "'^ 9 • Definition. If a person is toj-eceive a prize on condi- tion of some event happening, the sum of money for which his chance might equitably be sold beforehand is called his expectation from the event. KULE V. The eoopectation from any event is obtained hy multiplying the sum to he realized on the- event happening, hy the chance that the event will happen. This rule may be illustrated as follows: Suppose a person holds five tickets in a lottery, where the whole number of tickets is twelve : and suppose there be only one prize, and let its value be one shilling. The person in question gains the prize, if it happen that one of his tickets be drawn. The chance of this event is 5^; therefore, according to the rule, the person's expectation is j2 of a shilling, or five-pence. And the correctness of this result may be immediately seen ; for we observe, that if the EXPECTATION. 149 person had bought all the twelve tickets he would have been certain of winning a shilHng, and, therefore, he might equit- ably have given a shilling for the twelve tickets; but all the tickets are of equal value, and are equally valuable whether the same man hold one or more. Hence, each of them is worth a penny, and, therefore, the five in question are worth five-pence (as long as it is unknown which is drawn). Five-pence, therefore, is the sum that might equit- ably have been given for the assigned person's chance, and, therefore, by the definition this is his expectation. Question 109. A bag contains a £5 note, a £10 note, and six pieces of blank paper. What is the expectation of a man who is allowed to draw out one piece of paper ? Answer. Since there are eight pieces of paper the pro- bability of his drawing the £5 note is g ; therefore, his ex- pectation from the chance of drawing this note is g of £5, or g of a pound. Similarly, his expectation from the chance 1 5 of drawing the £10 note is g of £10, or | of a pound. There- 15 fore his whole expectation is -g- of a pound, or £1. 17s. Qd. Question 110. What is the expectation of drawing a coin from a bag which contains one sovereign and seven shillings ? Answer. The expectation from the chance of drawing the sovereign is g of a sovereign, and the expectation from the chance of drawing a shilling is g of a shilling. Hence, the whole expectation is 3s. 4Jd 150 CHANCE. Question 111. A person is allowed to draw two coins from a purse containing four sovereigns and four shillings. What is the value of his expectation 1 8 7 Answer. Two coins can be drawn in ^-0 '^^ ^^ ways: of these :r^ or 6 ways will give two sovereigns, 4 x 4 or 16 ways will give a sovereign and a shilling, and the other 6 ways will give two shilHngs. Therefore — Chance of drawing 40 shillings = ^ , Chance of drawing 21 shillings = gg , Chance of drawing 2 shillings = gg • Therefore the expectation is — from the first chance, ^r^ x 40, or -=- shillings ; 1 R from the second chance, jj^ x 21, or 12 shillings ; 6 3 from the third chance, hk x 2, or = shillings. Hence the whole expectation is -.^ + 12 + = , or 21 shillings ; or one-fourth of the whole sum in the bag. This result might have been inferred at once from the consideration that, if all the eight coins had been drawn twa and two, no drawing could be more likely to exceed in sovereigns than in shillings : (the number of sovereigns and shillings being the same). Hence the expectation from each of the four drawings must be the same ; and therefore each must be one-fourth of the whole sum to be drawn. CONCUBEENT EVENTS. 151' EULE VI. The chance of two independent events both happening, is the product of the chances of their happening severally. 5 That is, if the chance of one event happening be jj , and the chance of another independent event happening be g , the chance that both events should happen is g x g or jg . This may be proved as follows : — The chance of the first event is the same as the chance of drawing white from a bag containing six baUs, of which five are white (Eule IV.). The chance of the second event is the same as the chance of drawing white fi:om a bag containing eight balls, of which seven are white. Therefore the chance that both events should happen is the same as the chance that both balls drawn should be white. But the first ball can be drawn in six ways, and the second in eight ways. Therefore {Choice, Eule I.), both can be drawn in. 6 x 8, or 48 ways. So the first can be white in five ways, and the second can be white in seven ways. Therefore both can be white in 5 X 7, or 35 ways. That is, the two balls can be drawn in forty-eight ways (all equally likely), and thirty-five of these ways will give double-white. Hence (Eule IV.) the chance of double-white 35 is jg , and therefore the chance of the two given events both , . . 35 happemng is jg . 152 CHANCE. And the same reasoning would apply if the numbers were any others. Hence the rule is true always. Example. Suppose it is estimated that the chance that 2 A can solve a certain problem is g , and the chance that B can solve it is j2 ! let us consider what is the chance of the problem being solved when they both try. The problem will be solved, unless they both fail. Now the chance that A fails is g : and the chance that B fails is j2 ■ Therefore the chance that both fail is 3 ^12'°'' 36' The chance that this should not be so, is , 7 29 1 or — 36' 36 This is, therefore, the chance that the problem gets solved. In the case just considered, four results were possible, viz.: — (1) That A and B should both succeed : (2) „ A should succeed and B fail : (3) „ A should fail and B succeed : (4) „ A and B should both fail. We may calculate the chance of these four events sepa- rately. Thus we have Chance of ^'s success = ^ , of A's failure = ^ ; o o „ B's success = r-^ , of B's failure = ^r^ . CONCUREENT EVENTS. 153 Therefore, by the rule, 2 5 10 (1) Chance that A and B both succeed = o ^ ts ~ oc ■ 2 7 14 (2) Chance that A succeeds and B fails == 5 ^ To — oS = 15 5 (3) Chance that A fails and B succeeds = s ^ To = oc ■ 17 7 (4) Chance that A and B both fail = ^ x TS = oc • We observe that 10 14 5 7^36^ 36 ■*■ 36 36 "^ 36 ~ 36 ' or the sum of the four probabilities is unity, as it ought to be, since it is certain that one of the four results must happen. Further, we notice that the problem will be solved if any of the first three events out of (1), (2), (3), and (4) occur. Hence the chance of the problem being solved might have been obtained by adding together the separate probabilities of these three events. Thus — 10 14 ^^29 36 "•" 36 ■'' 36 ~ 36 ' 29 or the probability is gg , as before. KULE VII. If there he two events which are not independent, the chance that they should both happen is the product of the 154 CHANCE. chance that the first should happen, and the chance that when the first has happened the second should happen also. This case differs from the case of Rule VI. by this cir- cumstance, — that the second event, instead of being quite independent of the first, is so related to it that its proba- bility is altered by the first event happening. For example, if two letters are drawn from an: alphabet of 20 consonants and six vowels, the chance of the second letter being a vowel, which was originally gg, becomes 25 ^'^ 25 ^-ccording as the first letter was a consonant or a vowel. If however we are to calculate the chance of both being vowels, we have nothing to do with the second drawing except in the case when the first letter was a vowel. Our chance of drawing two vowels is plainly the same as if we had first to draw one from an alphabet of 20 consonants and six vowels, and then indepen- dently to draw one from another alphabet of 20 consonants and five vowels. Thus it appears that when the chance of the second event is dependent on the issue of the first, we may treat them as two independent events, if only we take for the second not its original probability but its probability when the first event has happened. Hence the truth of Rule VII. is manifest. The result in the particular case just considered, viz., that the chance of drawing two vowels out of an alphabet of 20 consonants and six vowels is 26 ^ 25 ~ 65 ' is verified by observing that as a drawing of two letters can be made in 26 . 25 ways, and a drawing of two vowels in 6 . 5 ways {Choice, Rule IV.), the chance of drawing two vowels must be (by Rule III.) TWO EVENTS NOT INDEPENDENT. 155 6.5 _ 3 26.25 65' the same result as is given by the present rule. Question 112. One purse contains five sovereigns and four shillings; another contains five sovereigns and three shillings. One purse is taken at random and a coiu drawn out. What is the chance that it be a sovereign ? Answer. The chance that the first purse be selected 1 2' is o, aad if it be selected, the chance that the coin be a 5 sovereign is g: hence the chance that the coin drawn be one of the sovereigns out of the first purse is 2^9'°'^I8- Similarly the chance that it be one of the sovereigns out of the second purse is 15 5 2^8'°"^ 16- Hence the whole chance of drawing a sovereign is 5 5 85 1 or 18 16 ' 144 ■ Question 113. What is the expectation from the draw- ing of the coin ia the last question ? 85 Answer. The chance that it is a sovereign is jg, and therefore the expectation from the chance of drawing a ■ 85 » , 1700 , .,,. sovereign is j^ of a pound, or ^^f shimngs. If the coiu drawn be not a sovereign, it must be a shilling, therefore the chance of drawing a shilling must be 156 CHANCE. ■'• ~ iii ' °^ iS (^^^^ !•)■ Hence the expectation from the 59 chance of drawing a shilling is jg of a shilling. Therefore the whole expectation from the drawing is 1700 59^ 1759 144 "'"144' °^ 144 shillings, or 12s. 2-^d. Question 114. What would have been the chance of drawing a sovereign if all the coins in the last case had been in one bag; and what would have been the expectation? Answer. There would have been ten sovereigns and seven shillings in the bag; therefore, the chance of drawing a sovereign would have been j^ , and the chance of drawing 7 a shilling js (Rule I.). The expectation would therefore have been 200 7 207 1^ + 17 °'W shillings, or 12s. 2^jd. The chance of drawing a sovereign is therefore in this case a little less, and the whole expectation very slightly less than in the former case. Question 115. What is the chance that in a year named at random Easter should fall on April 25 of the Gregorian Kalendar ? Answer. By reference to the tables at the beginning of the Prayer Book it is seen that this event will occur if the golden number be that affixed to April 18, and if that day be Sunday. The golden number required occurs once in EXAMPLES. 157 nineteen years, and when it occurs the day is equally likely to be any of the seven days of the week. Therefore the re- quired chance is 111 X .* 19 7 133 Question 116. There are three parcels of books in another room, and a particular book is in one of them. The odds that it is in one particular parcel are three to two; but if not in that parcel, it is equally likely to be in either of the others. If I send for this parcel, giving a description of it, and the odds that I get the one I describe are two to one, what is my chance of getting the book ? Answer. The chance of getting the parcel described 2 .... 3 is g, and the chance that the book is in it is g; therefore the chance of getting the book in the described parcel is 2 3 6 The chance of getting a parcel not described is ^ , and the chance that the book is in it is g; therefore the chance of getting the book in a parcel not described is g x g , or jg • * Hence on an average Easter will be on April 25 about three times in four centuries. At present, however, and for more than 3000 years to come, it happens about once a century, at intervals of either 57, 68, 84, 95, 152 or 163 years. This excess of frequency is compensated for after the year 4900, when we have three intervals of 220, 1863, and 288 years respectively. The years in which Easter occurs so late as April 25, are 1666 1734 1886 1943 2038 2190 2258 2326 2410 2573 2630 2782 2877 2945 3002 3097 3154 3249 3306 3469 3587 3621 3784 3841 3993 4088 4156 4224 4376 4528 4680 4748 4900 5120 6483 6771 6855 &a. 158 CHANCE. Therefore the whole chance of getting the book at all is 6 17 . . jg + jg , or jg ; or the odds are eight to seven against getting it. Question 117. In a purse are ten coins, of which nine are shillings and one is a sovereign ; in another are ten coins, all of which are shillings. Nine coins are taken out of the former purse and put into the latter, and then nine coins are taken from the latter and put into the former. A person may now take whichever purse he pleases ; which should he select ? Answer. Since each purse contains the same number of coins, he ought to choose that which is the more likely to contain the sovereign. Now the sovereign can only be in the second bag, provided both the following events have taken place, viz. — (1) That the sovereign was among the nine coins taken out of the first bag and put into the second. (2) That it was not among the nine coins taken out of the second bag and put into the first. 9 Now the chance of (1) is jg, and when (1) has hap- pened the chance of (2) is jg ; therefore the chance of both ■9 10 9 happening is J5 x jg > oi" jg . This, therefore, is the chance that the sovereign is in the second bag, and therefore (Rule I.) 9 10 the chance that it is in the first is 1 — jq or js . Hence, the first bag ought to be chosen in preference to the other. SEKIES OF CONCURRENT EVENTS. 159 EULE VIII. The chance that a series of events should all happen is the continued product of the chance that the first should happen, the chance that {when it has happened!) then the second should happen, the chance that then the third should happen, and so on. This is a simple extension of the last rule. For suppose there be four events, and let ^ be the chance that the first should happen, and when the first has happened, let j be the chance that the second should happen, and when these have 5 happened, let g be the chance that the third should happen, and when, these have happened, let j be the chance that the fourth should happen; by Rule VII., the chance that the first and second should both happen is g x t , or g . We may now treat this as a single event, and then, again applying 3 5 15 the same rule, we get g x g , or gj as the chance that the first, second, and third should all happen. Treating this compound event as one event, we can again apply the same rule, and obtain gj x j , or 255 as the chance that all the four events should happen. Thus the chance of all the events is 1 3 5 .1 the continued product of all the given chances. Question 118. There are three independent events whose 231 several chances are g , g , g . What is the chance that one of them at least will happen ? 160 CHANCE. Answer. One at least will happen, unless all fail. 12 1 1 The chance of all failing is ^ x r x s > or — . 1 14 Hence the required chance is 1 — =-= , or^v • Question 119. There are three independent events whose 2 3 1 several chances are g, ?, g. What is the chance that exactly one of them should happen ? Answer. The chance that the first should happen and the others fail is 2 2 1 4 3^ 5 ^2' °^m- So the chance that the second should happen and the others fail is 3 11 3 5 ^3 "^2' °^W And the chance that the third should happen and the others fail is 112 2 2 ^3 "^5' ''''so- Hence, the chance that one of these should occur — that is, that exactly one of the three events should happen — is 4^ 3 2 d_ 3 30 "^30 "^30' °^30' """lO" Question 120. When six coins are tossed, what is the chance that one, and only one, will turn up head ? Answer. The chance that the first should turn up head is 2) and the chance that the others should turn up tail is EXAMPLES. 161 ^ for each of them. Therefore, the chance that the first should turn up head and the rest tail is 111111 1 And there will be a similar chance that the second should alone turn up head, or that the third should alone turn up head, and so on. Hence, the whole chance of some one, and only one, turn- ing up head is A J_ Jl Jl J_ J_ A 64 ■*■ 64 "•■ 64 "•" 64 "^ 64 "^ 64 ' °^ 64* Question 121. When six coins are tossed, what is the chance that at least one will turn up head ? Answer. The chance that all should turn up tail is 111111 1 2^2^2^2^2^2' °'"64' The chance that this should not be so, or that at least one head should turn up, is (Rule I.) , 1 63 ^-64' ""^64- Question 122. A person throws three dice, what are the respective chances that they should fall all alike, only two alike, or all different ? Answer. The chance- that the second should fall the same as the first is g , and the chance that the third should also fall the same is g . Hence, the chance that all three fall alike is 11 1 6'^6'*''36- w. 11 162 . CHANCE. The chance that the second should fall as the first, and that the third should fall different, is 15 5 6^6' """"M' and there is the same chance that the second and third should be alike, and the first different ; or that the first and third should be alike, and the second different. Hence, the chance that some two should be alike, and the others differ- ent, is 36 "^36"*" 36' ""^Se" The chance that the second should be different from the first is y , and the chance that the third should be different 4 from either is g . Hence, the chance that all three are differ- ent is 5 4 20 6^6'°^ 36- Therefore, the three chances required are 36« 36> Ic ^^^pec- tively, their sum being unity, since the dice must certainly fall in some one of the three ways. Question 123. A person throws three dice, and is to receive six shillings if they all turn up alike, four shillings if two only turn up alike, and three shillings if all turn up' different, what is his expectation ? Answer. Referring to the last question, the chance of all turning up different is ^ ; his expectation from this event is therefore ^ of six shillings, or two pence. The chance of two only turning up alike is g^ or j^ , and his expectation EXAMPLES. ] 63 5 from this event is therefore j^ of four shillings, or twenty pence. The chance of all turning up different is gg or g , and his expectation from this event is therefore g of three shil- lings, or twenty pence. Therefore his whole expectation is 2 + 20 + 20, or 42 pence, or three shillings and sixpence. Question 124 A person goes on throwing a single die until it turns up ace. What is the chance (1) that he will have to make at least ten throws; (2) that he will have to make exactly ten throws ? Answer. (1) The chance that he fails at any particular 5 trial to throw an ace is - . The chance that he should fail 6 the first nine times (by Rule VIII.) is (^) . This, therefore, is the probability that he will have to throw at least ten times. (2) Since l-r) is the chance that he fails the first nine times, and 77 the chance that he succeeds the next time, o therefore by Rule VII., (-| x ^ is the chance that he will have to throw exactly ten times. Question 125. A die is to be thrown once by each of four persons. A, B, C, D, in order, and the first of them who throws an ace is to receive a prize. Find their respective chances, and the chance that the prize will not be won at all. Answer. Since A has the first throw, he wins if he throws an ace; his chance is therefore g . So B wins provided A fails and he succeeds. The chance 11—2 164 CHANCE. 5 1 of A failing is g , and of B succeeding is ^ . Therefoa-e B's chance of winning is 5 1 5 So G wins provided A and B both fail, and he succeeds. ... 5 5 25 ' The chance of A and B both failing is g x g , or gg ; and then the chance of G succeeding is g . Therefore G's chance of 25 1 ^ 36 ^6'°'' 216- So jD wins provided A, B, and all fail, and he succeeds. The chance of A, B, and G all failing is g x g x g , or gjg j and then the chance of B succeeding is g . Therefore B'& chance of winning is 125 1 125 X •^, or 216 6 ' 1296 ■ The prize is not won at all, provided all four fail to throw an ace. The chance that this should be the case is 5 5 5 5 625 X t: X TT X t; , or 6 6 6 6' 1296 ' Question 126. Two persons, A and B, throw alternately with a single die, and he who first throws an ace is to receive a prize of £1. What are their respective expectations ? A-nswer. The chance that the prize should be won at the first throw, is ^ : 5 1 at the second throw, is 7: x 7; : D o ^j X ^: EXAMPLES. 165 at the fourth throw, is •(^) x 6" /5\* 1 at the fifth throw, is I^J x^: 1 6" /SX" 1 at the sixth throw, is i^J x^ and so on. But the first, third, and fifth, &c., throws belong to A, and the second, fourth, sixth, &c., belong to B. Hence A's chance of winning is 1 , fBV 1 /5\* 1 » 6 + (6}-6 + (6)-6 + ^«-' and B's chance is 5 1 /5\' 1 /5\= 1 , 5 that is, B's chance is equal to ^'s multiplied by g . Hence 5 B's expectation is g of A's, or B's is to A's in the ratio of five to six. But their expectations must together amount to £1. Hence ^'s expectation is jj of a pound, and jB's j| of a pound. Question 127. What is the chance that a person with two dice will throw aces exactly four times in six trials ? Answer. The chance of throwing aces at any particular 1 ... 35 trial is ^ , and the chance of failing is gg . Hence the chance of succeeding at four assigned trials and failing at the other two is ( s^ ) ^ ( oe I • ^^^ ^^^^ ''^i^l ^^ thrown exactly four times if they be thrown at any set of four trials which might be assigned out of the six trials, and if they fail at the re- maining two. And (Choice, Rule IX.) it is possible to assign 166 CHANCE. 6 5 4 3 four out of six in ' ' " , or fifteen ways. Hence the JL . ^ • o ■ 4} chance required is fifteen times the chance of succeeding in four assigned trials, and failing at the other two. Therefore it is ^125 (^y^(i)'^i^' orT 725594112' therefore the odds are more than 100,000 to 1 against the event. Question 128. If on an average nine ships out of ten return safe to port, what is the chance that out of five ships expected, at least three will arrive ? Answer. The chance that any particular ship returns is 9 zp^ . The chance that any particular set of three ships should . /9 \* all arrive is ' :j-j; j , and the chance that the other two should not arrive is [^) ■ Therefore the chance that a particular ' / 9 \' / 1 \^ 729 set of three should alone arrive is I ^7.1 x (— -j , or . 5 4 3 And out of five ships a set of three can be selected in ' " or 10 ways. Hence the chance that some one of these sets of three should alone arrive is 729 ^- 729 X 10, or 100000 ' 10000" This is therefore the chance that exactly three ships should arrive. Similarly the chance that any particular set of four should alone arrive is (jA ^ in ' °' IO6OOO ' ^^^ *^® chance EXAMPLES. 167 that some one of the five possible , sets of four should alone chance that exactly four ships should arrive. And the chance that all the five should arrive is (^7;) , 59049 100000 ■ But the chance that at least three should arrive is the chance that either three exactly, or four exactly, or five exactly should arrive: and is therefore the sum of the several chances of these exact numbers arriving : that is, the required chance is 7290 32805 59049 99144 12.393 100000 "*■ 100000 "^ iOOOOO ' °'^' 100000 ' ""^ 12500 ■ Question 129. A and B play at a game which cannot be drawn, and on an average A wins three games out of five. What is the chance that A should win at least three games out of the first five ? Answer. The chance that A wins three assigned games, /^\^ /2\^ 108 and B the other two, is (^) . ( ^ ) , or -,,-. . But the three V5/ V»/ 312o 5.4.3 may be assigned in -.— g— „, or 10 ways {Choice, Rule IX.). Hence the chance that A should win some three games and B the other two is 108 , „ 1080 X 10, or , 3125 ' 3125 ■ Similarly the chance that A should win some four games and B the other one is 162 810 3125^ '*"^ 3125' 168 CHANCE. Arid the chance that A should win all five games is /3\' 243 UJ'^'siIB- Therefore the chance that A wins either three, or four, or all out of the first five games is 1080 810 243 ^2133 3125 ■'' 3125 3125 ~ 3125 ' or the odds are rather more than two to one in ^'s favour. EXPECTATION OF LIFE. The next four questions are based on some results of Vital Statistics. It will be assumed that out of 400 persons horsy, 57 die in their first year, 20 in their second, 11 in their third, and so on, according to the subjoined table, until at the age of 90, only 4 survive. A D A D A D A D A D A D A D A D A D 1 57 11 21 2 31 2 41 3 51 4 61 5 71 '7 81 5 2 20 12 22 2 32 2 42 3 52 4 62 5 72 7 82 5 3 11 13 23 2 33 2 43 3 53 4 63 5 78 7 83 4 4 7 14 24 2 34 2 44 3 54 4 64 6 74 7 84 4 5 5 15 25 2 35 3 45 3 55 4 65 6 75 6 85 4 6 3 16 26 2 86 3 46 3 56 4 66 6 76 6 86 3 7 2 17 27 2 37 3 47 3 57: 4 67 6 77 6 87 3 8 2 18 28 2 88 3 48 4 58 4 68 6 78 6 88 2 9 2 19 2 29 2 39 3 49 4 59 5 69 6 79 6 89 2 10 1 20 2 30 2 40 3 50 4 60 5 70 6 80 •5 90 1 110 12 20 26 33 42 57 ^ 63 33 EXPECTATION 01" LIFE. 169 And we shall further assume that of 100 persons arriving at the age of 90, 19 die in the next year, 17 in the next, 15 in the next, and so on in arithmetical progression, until at the end of the 100th year all are dead. The following a,pproximate results of the table are easily remembered. Out of 400 persons born 100 die before the age of 5 years, 100 more (accurately 101) die before the age of 50, 100 more (accurately 99) die before the age of 70, 100 survive 70 years. Again : 120 (accurately 122) die in the first 20 years, 120 (accurately 121) die in the next 40 years, 120 die in the next 20 years, leaving 40 (accurately 37) to survive the age of 80. Question 130. What is the chance of a person aged 30 living till he is 60 ? Answer. Out of 400 persons bom simultaneously with him 142 die in the first 30 years, and 258 survive. Of these 101 die in the next 30 years and 157 survive. The chance that the man in question is one of the survivors is therefore ^ . Question 131. What are the odds against a person aged 40 living for 30 years ? Answer. Out of 232 persons surviving to the age of 40 there will be 132 deaths in the next 30 years, and 100 will then survive. The odds are therefore 132 to 100 or 33 to 25 against the person living to seventy. ■ Question 132. What is the expectation of life of a person on his 90th birthday ? 170 CHANCE. '100 Answer. The chance that he dies the first year is jqq , the 17 second year jTyj , and so on, in arithmetical progression. But if he dies the first year (since he may equally die in any part of it) his expectation of life is only g year. So if he die in the second year his expectation is g years. If he die in the 6 third year it is -^ years, and so on. Hence Tiis whole expec- tation is 19 . 1 + 17 . 3 + 15 ■ 5 + 13 ■ 7 + ... + 3 . 17 + 1 ■ 19 200 7 which will be found to amount to 32^ . Question 133. What is the e'xpectation of life of a person on his 81st birthday ? The chance that he dies the first year is -^ , the second, 4 . 3 third and fourth each gg , the fifth and sixth each ^ , the 2 .1 seventh and eighth each ^ , the ninth ^ , and the chance that he survives this, i.e., that he reaches his 90th year, is 55. His expectation of life prior to the completion of the 90th year is therefore 5 . 1 + 4 (3 + 5 + 7) + 3 (9 + 11) + 2 (13 + 15) + 17 99 64 32' And his expectation subsequent to his 9Qth year is 32(9 + 325) =jgQ, the nine years being added to the result of the previous question. Hence his whole expectation is 99 247 _ 742 _ 51 32'''160~160~ 80' ACCESSION OF KNOWLEDGE. 171 RULE IX. If a doubtful event can happen in a number of different ways, any accession of knowledge concerning the event which changes the probability, of its happening will change, in the same ratio, the probability of any particular way of its hap- pening. It follows ^om the -axiom that the probability of the event happening at all must be equal to the sum of the probabilities of its happening in the several ways. First, suppose for simplicity that all the ways are equally likely. Let there be seven ways, and let the chance of each one severally occurring be jq : then the chance of the event 7 happening at all is seven times this, or jq . But suppose that our knowledge is increased by the infor- mation that the event happens nine times out of ten, or by such other information as brings our estimate of its probability 9 . 7 , . ... up to jg instead of jq , thus increasing the probability in the ratio of seven to nine. It is still true that there are only seven ways of the event happening, all of which are equally likely : hence the proba- bility of the event happening in any particular one of these 19 9 ways is y of jq , or =q , with our new information. Hence our information concerning the event has increased the chance 17 9 of its happening in an assigned way from jq ''^ to ^° 70 ' *^** is, it has increased it in the ratio of seven to nine, the same ratio in which the probability of the event itself was in- creased. * 172 CHANCE. And the same argument would hold if the numbers were any others, and therefore the rule is true, provided all the ways of the event happening are equally probable. Secondly, suppose the ways are not equally probable. We may in this case regard them as groups of subsidiary ways, which would be equally probable. Then, as we have shewn, the chance of each one of the subsidiary ways would be increased (or decreased) in the same ratio as the chance of the event itself, and therefore the sum of the chances of any group of these subsidiary ways would be changed in the same ratio. For instance, if the event could happen in any one of three ways, whose respective chances were 3) g' i' ^"^ 12' 12» 3 . . . J2, we might divide the first of these ways into four sub- sidiary ways, the next into two ways, and the other into three ways, and the chance of each of these subsidiary ways would be j2- If therefore, by an accession of knowledge, the chance of the whole event were diminished in the ratio of three to two, each subsidiary way of the event's happening 2 1 1 would have a diminished probability of » of js > or jo , and .the probabilities of the three given ways would become 423 211 respectively jg, j^, jg, or g, g, g: that is, they would be diminished in the same ratio as the chance of the event itself. Thus we see that the rule is true always. Question 134. A bag contains five balls, which are known to be either all black or all white — ^and both these are equally probable. A white ball is dropped into the bag, and then a ball is drawn out at random and found to be EXAMPLES. 173 white. What is now the chance that the original balls were all white ? Answer. The prohabilities are here affected by the ob- served event that a ball drawn out at random proved to be white. We will first calculate the probabilities before this event was observed (which we will call m priori probabilities), and then consider how they are affected by the accession of knowledge produced by the observation of the event. (Pro- babilities modified by this knowledge may be distinguished as a posteriori probabilities.) The event might happen in two ways; either by the balls having been all white, and any one of them being drawn, or by the five original balls having been black, the new one alone white, and this one drawn. The a priori probability that all are white is g, and then the chance of drawing a white ball is 1 (or certainty). Hence the chance of the event happening in this way is 2Xl,or2. So the a priori probability that the first five were black is g , and then the chance of drawing a white ball is g. Hence the chance of the event happening in this way is 11 i_ 2 X g , or j^2 • Therefore the whole a priori chance of the event happen- . -1,1 7 ingiS2 + i2.orj2- But when the ball is drawn and observed to be white, 7 this knowledge immediately increases the chance from th to 1 (or certainty) : that is, it increases the chance in the ratio of 7 to 12. Therefore, by Rule IX.,' the chances of the event 174 CHANCE. happening in the several ways are increased in the same ratio. Hence the d posteriori chance of the event having hap- pened in the first way is g x y , or y ; and the A posteriori 1 12 chance of its having happened in the second way is j2 ^ T ' or s. Or the chance of the original balls having been all white is now ^, and of their having been all black =. Question 135. In a parcel of 1000 dice there is one that has every face marked six : all the rest are correctly marked. A die taken at random out of the parcel is thrown four times and always turns up six. What is the chance that this is the false die ? Answer. The die is either false or true : and the respec- tive chances of these two cases a priori are j^gg and j^qq . In the first case the chance that six should turn up four times in succession is unity; in the second case it is (gj , or J296 • Therefore we have a priori, — (1) the chance that the die should be false, and dx should turn up four times, is jJqq : (2) the chance that the die should be true, and six 37 should turn up four times, is iggoo '■ and the chance that from one or other the same result should happen is 1 37 17 1000 ' 48000 9600' But after our knowledge is augmented by the observation of the fact that the die turns up six four iimes in succession. EXAMPLES. 175 this latter chance becomes unity, that is, it becomes multi- plied by -js- . Hence (Rule IX.) the chances (1) and (2) become multiplied in the same ratio. Therefore a posteriori, the chance that the die should be false and six have turned up four times is jqqq x -^r — gg . This is the required chance. Question 136. A purse contains ten coins, each of which is either a sovereign or a shilling : a coin is drawn and found to be a sovereign, what is the chance that this is the only sovereign ? Answer. A priori, the coin drawn was equally likely to be a sovereign or a shilling*, therefore the chance of its being a sovereign was g . A posteriori, the chance of its being a sovereign is unity ; or the chance is doubled by the observation of the event. Therefore (Rule IX.) the chance of any particular way in which a sovereign might be drawn is also doubled. Now the chance that there was only one sovereign was a priori 10 J^ 2" °'' 1024' and in this case the chance of drawing a sovereign was 1 lo- * In the statement of this qnestion the words "each of which" implies that the purse has been filled in such a way that each coin separately is equally likely to be a sovereign or a shiUing. For instance each coin may have been taken from either of two bags at random, one containing sovereigns and the other shiUlugs. The case is carefully marked off from that of Qn. 137. Mr Yenn in his strictures on this solution (Logic of Chance, Second Edition, pp. 166, 167) appears to overlook the significance of the words "each of which," and implies that the solution of Qn. 137 would have been appli- cable to Qn. 136. 176 CHANCE. Hence the chance that there should be only one sovereign, and that it should be drawn, was d, priori 10 1 1 • X tt; , or : 1024 10 ' 1024 ■ And the A posteriori chance that a sovereign should be 2 1 drawn in this way is the double of this : i.e. j^j or gjg ; which is therefore the required chance., Question 137. A purse contains ten coins, which are either sovereigns or shillings, and all possible numbers of each are eqiially likely : a coin is drawn and found to be a sovereign, what is the chance that this, is the only sovereign.? Answer. A priori, the coin drawn was equally likely to be a sovereign or a shilling, therefore the chance of its being a sovereign was ^ . A posteriori, the chance of its being a sovereign is unity :■ or the chance is doubled by the observation of the event. Therefore (Rule IX.) the chance of any particular way in which a sovereign might be drawn is also doubled. Now, a priori, eleven cases were equally probable, viz. that there should be 0, 1,, 2,, 3, 4, 5, 6, 7, 8, 9, 10 sovereigns, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, shillings. Therefore the chance of there being exactly one sovereign was jj , and in this case the chance of drawing a sovereign Hence the chance that there should be only one sove- reign, and that it should be drawn, was, d priori, 11 "^10' """no- EXAMPLES. 177 And the a posteriori chance that a sovereign should be 2 1 drawn in this way is the double of this, that is, no ""^ 55' which is, therefore, the chance required. Question 138. A, B, G were entered for a race, and their 2 4 5 respective chances of, winning were estimated at U' il' li" But circumstances come to our knowledge in favour of A, which raise his chance to g; what are now the chances in favour of B and C respectively ? Answer. A could lose in two ways, viz. either by B winning or by C winning, and, the respective chances of his 4 5 losing in these ways were d priori jj and jj , and the chance of his losing at all was tt . But after our accession of know- ledge the chance of his losing at all becomes ^i tli^t is, it becomes diminished in the ratio of 18 : 11. Hence the chance of either way in which he might lose is diminished in the same ratio. Therefore the chance of B winning is and of winning 4 11 £ ll''l8' '''' 18' ^11 ^ 11^18' °'' 18- These are therefore the required chances. Question 139. One of a pack of fifty-two cards has been removed; from the remainder of the pack two cards are drawn and are found to be spades ; find the chance that the missing card is a spade. Answer. A priori, the chance of the missing card being w. 12 178 CHANCE. a spade is j , and the chance that then two cards drawn at 12 11 132 random should be both spades is sj-gg > or gggg . Hence the chance that the missing card should be a spade, and two spades be drawn, is 1 132 11 4 2550' 850* 3 The chance of the missing card being not a spade is j , and 13 . 12 the chance that then two spades should be drawn is 53—50 , or 156 2ggQ. Hence the chance that the missing card should be not a spade, and two spades be drawn, is 3 156 39^ 4^2550* °^ 850" Therefore the chance that in one way or the other two spades should be drawn is n^ 39^ ^ 1^ 850 "^850' °^ 850' °^ 17" But after the observation of the event this chance be- comes certainty, or becomes multiplied by 17. Therefore the chance of either way from which the result might occur is increased in the same ratio. So the chance that the given card was a spade becomes a posteriori, 850^1^'°^ 50- Question 140. There are four dice, two of which are true and two are so loaded that with either the chance of throwing six is 3 . Two of them at random are thrown and turn up CREDIBILITY OF TESTIMONY. 179 sixes. Find the chance (a) that both are loaded; (b) that one only is loaded ; (c) that neither is loaded. Answer. The &c- NOTES ON POPULATION, AND EXPECTATION OF LIFE. 1. Let Sj, Sj, Sj . . . be the respective chances that a person die in the 1st, 2nd, 3rd . . . year of his life. The arithmetical values of these constants have been already given in the table on page 168, the entries in Column D being the numerators to denominator 400. That is, s_il s-^ S-il S-JL* '~400' =" 400' » 400' ^~400' 2. Out of iV" persons born in a given place in the course of a year, let 0„N be the number who die in the same year, 6jJV the number who die in the next year, 0Jif in the next, and so on. It is plain that if all the births took place at the first moment of the year we should have And if all the births took place in the last moment of the year we should have NOTES ON POPULATION. 195 The births being actually spread over the whole year, we take as a sufficient approximation the mean between the fore- going. Hence and so on. The values of S being known from the table on page 168, the values of are now known. 3. Excluding the consideration of immigration and emi- gration, if B be the number of births in a year, and D the number of deaths in a year, in a population which numbered iV" at the beginning of the year, the population at the end of the year will number N + B — D. In other words the population :will be multiplied in a year by , B D '^N~N^'^ suppose. We may take the births and deaths in a year to be propor- tional to the population at the beginning of the year, so that /i may be treated as a constant. 4. Since the population is multipUed every year by im, if B and D be the respective numbers of births and deaths in B D any year, — and — will be the numbers in the previous year, r* ■ r' B D -j and -5 in the year before that, and so on. The deaths D in a given year must be composed of the following: 6^ B deaths of persons born in the year, 6^ — „ „ „ the previous year, 6^ -2 „ „ „ the year before that, r' 13—2 196 CHANCE, and so on. Hence D = B 1^ F F ) 2 l/i /i' /i' ) But (Art. 3) ^ = |-(^_1). . B ii-l ■^ i_/* + lf«t^«.^S, 2 [jj, ' fj,^ ' fj,^ Thus we have two equations connecting the birth-rate, the death-rate, and the. annual multiplier of population, so that when any one of these three is given, the other two are determinate. 5. Let iij, J?2, -Bj . . . represent the remainders of the series 81 + 82 + 83+... after 1, 2, 3, . . . terms respectively have been taken, so that B^ = B^^^ + S^ + S^^ + . . . . Then out of N' persons bom, NR„ will be the number who survive the age of X years; and the chance of a person aged m years living for at least n years more will be i2^^ -7- i?„. 6. Observing that B^ = iZ^_i — B^, and i?„ = 1, the last equation of Art. 4 becomes B 2/* N' -fr+^)if+f+f+-r 7. Let Eg, E^, E^... be the expectations of life of persons aged 0, 1, 2, ... years respectively. We may suppose the deaths in each year to be equally distributed over the year so that if a man is to die in his first year his expectation of life is half a year; if he is to die in his second year his expec- tation is a year and a half and so on. Then EXPECTATION OF LIFE. 197 for a man aged x years has tte chance S^^, -f- B,^ of dying in his first year, and the complementary chance of surviving his first year. In the one event he has an expectation \, and in the other an expectation E^^ dating from a year hence, or 1 + E^^ from the present date. Since S^, =Rx~ -K^+i the last equation reduces to a formula by which it is easy to calculate successive values of^. 8. To find the chance that M who is aged, m years, will survive N who is aged n years. The chance that M should die in a year subsequent to iV" is easily seen to be and the chance that they die in the same year is In the latter case the chance that M survives i\/' is ^. Therefore the whole chance of M surviving N is Or (observing that 2B^^^ + S^,^^ = R^ + R^^^, and that 7)1 n which is 2J2„i?. CHAPTER VIII. SUCCESSIVE TRIALS. If it be our object to give accuracy to popular estimates of chance and expectations in cases when the event is capable of experimental repetition, we shall be best understood if we measure the probability of the event by defining the number of times that the experiment would have to be repeated on an average before success is reached. In the case of a simple event of which the chance of success at any experiment is always the same, the average number of trials required for success is the reciprocal of the chance of succeeding at any trial. If it be the question of throwing doublets with a pair of dice, the chance of success at any trial is ^, and, on an average, 6 trials will be required in order to succeed. Of course the operator may succeed at the first trial, or he may fail for 6 or 60 or 600 trials ; but if the player go on indefinitely always counting the number of the throws he has had to make in trying to throw doublets he will find that on an average six throws were wanted to attain success. And the statement that an average of six throws will be required in order to throw doublets must not be confounded with the statement of the number of throws required in order that success may be more likely than failure. As a matter of fact if the operator make only four throws he is more likely EXPECTATION OF SUCCESS. 199. to throw doublets than not, (the odds being 671 to 625 in favour of success) ; and yet it is obvious that in the long run success will occur once in six trials, and that in an indefinite period of play, there will be on an average six throws for every success. In questions in which the chance of success at any experi- ment is not always the same, but depends (for instance) on the result of the previous trial, the average number of trials required for success is no longer the reciprocal of the chance of succeeding at the first trial. To this class will belong questions respecting sequences of results. For example : if it be required to throw doublets twice in succession the chance of succeeding immediately is gg , but the average number of throws required for success is 42 (Infra Prop. LIV.). So the odds against throwing sixes three times consecutively are 46655 to 1, but the average number of throws required for success will be 47988. The rarity of long sequences of heads or tails in tossing a coin cannot be better exhibited than in the statement that on an average we shall have to toss 2 (2" — 1) times to get a sequence of n heads. Thus if we make 10 tosses per minute we must expect to work 13 hours and 39 minutes to get a sequence of 12 heads; and 218 hours and 27 minutes to get a sequence of 16 heads. The term Expectation is usually limited to cases in which a person is to receive a sum of money contingent on the issue of some doubtful event. (See the definition, page 148.) Thus if a showman receives a penny for every shot at " Aunt Sally", and gives a cocoanut for every successful shot, and if a player succeeds once in seven shots, the showman's expec- tation as a return for each cocoanut is sevenpence : and the player's expectation is the value of a cocoanut minus seven-- 200 CHANCE. pence. But we may well speak of expectation independently of money, and say that the player has an expectation of seven shots. Similarly if a man tosses a coin till he gets a sequence of 4 heads we may say that his expectation is 30 tosses. PROPOSITION XLIX. If on an average success he attained in s trials, then will where f denotes the chance of failing in the first r trials. For if /tj be the chance of succeeding at the first trial ; and after the first has failed, fi^ be then the chance of suc- ceeding at the second trial; and after the first and second have both failed, fi^ be then the chance of succeeding at the third trial, and so on, we shall have (Prop. XLIV. Coi-. 2) s = Mi + 2/;/*, + 3/^3+4/>,+ .... f But /i^ = 1 — ^— for all values of r : therefore s = l+2f + 3f + if, + ...-{f+2f,+ Sf + ...) = l+/i+/.+/3+-- Q-E.D. Note. The series will be continued to infinity unless in the nature of the case success within (say) r trials is certain. In this ca.se f,f^^,... will each be zero. Corollary. If the chance of failing at any trial be /, then and s = 1 +/+r +/' + •■• to infinity _ 1 ^1 1-/ H-' where (i is the chance of succeeding at any trial. AVERAGE. 201 The result ia this particular case might have been antici- pated. It simply expresses the fact that if the chance of success is - , success will occur on an average once in n trials. Example. To draw double-blank from a set of 28 domi- noes will require on an average 28 trials if the dominoes drawn be replaced, but only an average of 14^ trials if they be not replaced. For by the Corollary, we have in the for- mer case 27\ :.I.(.-i) = .S, and in the latter, by the Proposition, , 27 , '26 25 ^ , -_ , 29 ^=^ + 28 + 28+^+-*° ^^*"™' = T- PEOPOSITION L. If an experiment succeed on an average once in n trials, the chance that it fails throughout n given trials is never greater than - ; and if n he large this chance is approxi- mately - . For the chance of failing in any trial is 1 , and the 1 — J . But by ex- pansion, \\ j is seen to be greater than e, and therefore f 1\" 1 . . . / W" 1 (1 ) < - ; and, as n is increased indefinitely, 1 1 ) = - . Therefore &c. Q. E. D. 202 CHANCE. Note. The value of the reciprocal of e is -367879 values of the chance are as follows : When n= 10, Chance =-34867, „ n= 20, „ = -35848, „ n= 30, „ = -36162, „ n= 50, „ =-36417, „ n= 100, „ = -36604, w = 1000, „ = -36771, The PKOPOSITION LI. If an event happen at random on an average once in time T t, the chance of its not happeniiig in a given period t is e"i . Divide each unit of time into an indefinitely large number n of small instants. Then in the periods t, r there will be nt, nr respectively of such instants. By making n large enough we may make the instants so short that the event may be said to happen at one such instant. Since it happens on an average once in nt instants, the chance of its happening at any particular instant is — , and the chance of its not hap- pening in the given period of wt instants is (-r=[(-irp- But when n is indefinitely large, II -\ ultimately = - . Therefore the required chance is (-)* , or e~7. Q. e. d. Example. If an earthquake happens on an average once CONSECUTIVE SUCCESSES. 203 a year, the chance that in a given year there should not be an earthquake is - . e PROPOSITION LII. If an experiment succeed m times and fail n times, the chance that among the m,+n trials there are nowhere k con- secutive successes is the coefficient of x^ in the expansion of For all the trials being equally likely to succeed all pes-" sible arrangements of the m successes and n failures must be equally likely to occur. Now the arrangements >vithout k consecutive successes may be made by regarding the n failures as standing in a row and then distributing into the n + 1 spaces between them (inclusive of the ends of the row) the m successes so that no parcel may have more than k — \. By Prop. XXVIII. (writing m for n and putting g' = 0, z = k, r = n + V) the number of arrangements thus made is the co- efficient of a;" in the expansion of \l-xj ■ And the total number of arrangements is \m-\-n-^\m \n. Whence the truth of the proposition is evident. PROPOSITION LIII. If fi-he the chance of an operation succeeding in any trial, the chance that in n trials there shall not he k consecutive suc- cesses is the coefficient ofx" in l-a;+(l-/i)/i'a;'+'' 204 CHANCE. For the chance that there should be exactly r successes among the n trials is \n \r \n — r And if there are r successes and n — r failures the chance that there should not be k consecutive successes is, by the last proposition, the coefficient of of in the expansion of \r\n — r 1 _ r* '"' ■ Z»-'-+\ where Z = 1-x Hence the chance of there being exactly r successes but not k consecutive successes is the coefficient of of in the expan- sion of ijr{l - (i)"''' X'"'"^\ which is the coefficient of ai" in the expansion of fjiT (1 - /i)"-'-^"-'' X"-'^\ Now the whole chance required will be obtained by giv- ing r all its possible values (from to m inclusive) and adding. Hence the chance will be the coefficient of as" in the expan- sion of And the series within the brackets may be continued to in- finity if we please, since the terms thus added contain only higher powers of x and cannot affect the coefficient of jc". Hence the chance required is the coefficient of «" in But the coefficient of a;" in any function is not altered if we CONSECUTIVE SUCCESSES. 205 write fix for x in the function, and then divide by /u.". There- fore the chance is the coefficient of x" in 1 — uV Note. By actual division, we find that the first k coeffi- cients are all unity, a result which corresponds to the fact that iine, or ('1 + — ) = 6/3 + C7 + ... = the sum of the prizes. Also let mD, ==b0' + cy^+ ... and let x be the price which the man in question ought to pay for the ticket. Then if we proceed as in the last Proposition we find the average multiplier for m ventures, viz. , \ n/ \ n nj \ n nJ This must be equal to unity. Therefore .loB(l-!)+M.g(l + f-|).„,og(l.^-!)+....„. PRICE TO BE PAID FOB A LOTTERY TICKET. 227 from which equation a; is to be found. If, as is usually the case, the amount at stake, is small compared with the specu- lator's whole funds, n must be a very large number ; there- fore we shall obtain an approximate value of x, by neg- lecting the terms involving high negative powers of n, in our equation. Thus, expanding the logarithms, we have . _ mx — riKo irw? — 2miox + mil whence x = a j^ — , zn which is therefore the price which a man with funds n may prudently give for the ticket whose mathematical value is a>. Corollary. The price which a man whose available fund is n pounds may prudently pay for a share in a specu- lation in which p^ will be his chance of winning P^, p^ his chance of winning P^, and so on (where P1+P2+ ■•■ = ^) will be Examples. Having n shillings, what may one prudently pay to be entitled to a number of shillings equal to the number turned'^p at a single throw of a die ? Here 2(pP) = g (1 + 2 + 3 + 4 + 5 + 6) = ^, 2(^0=^(1+4 + 9 + 16 + 25+36) = ^. 15—2 228 CHANCE. And the required number of shillings is approximately 7/, 5 Ml-— V 2 V 12Ji/ Again. If the throw be made with two dice, we shall have (see page 146) 2 {pP) = 7, t ipP"), = 329 -=- 6. And the required number of shillings will be approxi- mately It must be remembered that the results of the two fore- going propositions and the coroUary are only obtained on the hypothesis that n is very large compared with 2 (pP) and with S (pP^) -^ S ipP)- This requires that the man's original fund should be very large compared not only with the .amount which he stakes, but also with the amount which he has a chance (though it may be a very small chance) of winning. In all practical cases the former conditions will be ful- filled, as no one would think of staking, in any single ven- ture, his whole property, or a sum bearing any considerable ratio to his whole property. But cases will be likely to arise in which the latter condition is not satisfied, as when a man may purchase, for a sum comparatively small com- pared with his means, a ticket in a lottery in which there is a prize very many times larger than his whole property. In this case, the approximate results obtained above cannot be applied, and we must have recourse to the original equa- tion in the form in which it was presented before we intro- duced the approximations which are now inadmissible. We PEICE TO BE PAID FOR A VENTUEE. 229 may, however, still approximate, in virtue of the condition that the stake must be small compared with the speculator's whole funds. Thus : — PROPOSITION LX. To find an approximate formula for the sum which a speculator may pay for any defined expectation, without assuming that his funds are necessarily large compared with the value of the prizes. Let X (small compared with n, though not necessarily small compared with P) be the sum which a man whose available fund is n may prudently pay for the chances Pi? Pi' ^3---of receiving prizes worth P^, P^, P,, ... respec- tively, as in the last corollary. Then the rigorous equation to determine X is P, X\i'i/, . P, X\V^f, P, X\P, \ n n/ \ n nJ \ n n j which may be written or, since X is small compared with n, = 1, whence X -1 j>i I i>. I P^ I n + Pjn+Pjn + P, 8 230 CHANCE. This formula is equally applicable when there is a pos- sibility of not receiving any prize, as the failure to receive a prize may be treated as the receiving of a prize of zero value : i.e. one of the quantities P„ P^, Pg ... will be in this case zero. COEOLLAEY. In the case when there is a single prize P, and the chance of gaining it is p, the formula becomes P\P Z = P ^3'- n+P ll ^ n{n+P) (/ ^Y.-A , 1-p n + (l-p)P\V-^n) ^r Examples. A man possessing a pound is offered a ticket in a lottery, in which there are 99 blanks, and one prize worth 100 pounds. What may he prudently pay for the ticket ? Here n = l, P = 100, P = jJq. X = 1^ aWl - 1) = ^ X -0472 = -0519. Hence he can only afford to pay about a shilling for the ticket. Again. How much may a man with 10 pounds pay for the same ticket ? Here n = 10, P = 100, P = ^q- „ 1100 (m,~ _. 1100 „„,„ „ X = -jQQ- ( Vll - 1) - 309- X -0242 = -244. Hence he can afford to pay about 4s. 10|d. for the ticket. PRICE TO BE PAID FOE A VENTURE. 231 Again. How much may a man with 100 pounds pay for the same ticket ? Here n=100, P = 100, P = ^- X=^V72-l)=?^x -0069 = -693. Hence he may pay nearly 14 shillings for the ticket. Again. Suppose the man has 1000 pounds. This sum is large compared with the value of the prize. Hence we may apply the simpler formula given in Proposition LIX. (Cor.), which leads to the result Hence he may pay 19 shillings for the ticket. If we had applied to this last case the formula of the present corollary we should have obtained the slightly more correct result "9507, the difference between the two results being nearly one-fifth of a farthing. The formula of Proposition LX. is applicable in the case of the Petersburg problem, a problem of some intrinsic in- terest, but chiefly of importance on account of its having been repeatedly made the ground of objections to the mathe- matical theory of probability. This celebrated problem may be stated as follows : (PROP. LXI.) THE PETERSBURG PROBLEM. A coin is tossed again and again until a head falls. The player is to receive a florin if head falls the first tim6: two 232 CHANCE. florins if it falls the second timie, and not before: four florins if it falls the third time, and not before : and so on, dovhling every time. To determine what the player might advantage- ously give for the expectation? The absolute value of the expectation is easily seen to be infinite. Thus — The chance of winning the florin at the first throw is J : the value of the expectation is therefore half a florin. The chance of failing at the first throw and winning the two florins at the second throw is ^: the value of the ex- pectation is therefore \ of two florins, or half a florin. So, the chance of winning the four florins at the third throw is ^ : the value of the expectation is therefore ^ of four florins, or half a florin, and so on. Hence the value of the expectation attaching to each throw to which there is a possibility of the play extending is half a florin. If the play were limited to 2n throws the expectation would be n florins. But the play may extend to a number of throws larger than any assignable number. Therefore the whole expectation is worth a number of florins larger than any assignable number : that is, it is infinitely great. I But here a difficulty is raised. The mathematical ex- pectation has been found to be of infinite value, and yet (it is objected) no one in his senses would give even such a moderate sum as £50 for the prospect defined in the problem. The fallacy of this objection has been pointed out already. A man who is playing for even limited stakes is at a dis- advantage from the risk of his limited means terminating his operations at an unfavourable epoch. Much more is he at a disadvantage when the interests at stake are (as in this THE PETERSBUEG PROBLEM. 233 case) very large or even infinite. The absolute value of a mathematical expectation is not the prize which a man of limited means ought to pay for the prospect. It expresses the value of the expectation to a man who is able to repeat the venture indefinitely without the risk of his operations being ever terminated by lack of means. The speculator's fund, to begin with, must be infinite in comparison with the stakes involved, before he may venture to give the absolute value of the mathematical expectation for any contingent prospect which he may desire to purchase. In the Peters- burg problem the mathematical expectation is infinite : but if one is to give an infinite sum for the venture one must take care to hold funds infinite in comparison with this in- finity. In other words, the speculation on . these terms is only proper for one with respect to whose funds the infinite stake is inconsiderable. The stake which he lays down may be CO , provided that his funds are oo x oo . But to find the sum which a man of limited means may pay for the expectation defined in our statement of the problem we may apply the result of Propositio'u LX. Thus if n be the number of florins which the man possesses, the formula to determine the sum which he may pay may be written as follows : X = 1 1 1 2(71+1) 4(to+2) 8(n+4) and observing that „Hi+H- ... = „, and that 2* . 4* . 8'^''^ . . . = 2*"'"*''"*'^ ■■• = 2, 234 CHANCE. we get n ^— r If n be any power of 2 we can put the result into a form free from infinite series. Thus, if and are constants, viz. 6I = (1 + 1)* fi +|?(l + \f ...ad inf. = 1-6253... and we shall have - (3^f (1 + 4)»(1 + 8)".,. (1 + n)4 - 2 1 1 1 1 ^' »i + I 2 {n + 2)'^ 4> (n + i) '^ ^n {n + ^n) '^ n" The result when m is given is easily evaluated by the aid of a table of logarithms. For instance, if n = 8, the numerator is 7342 and the denominator is '1934. Hence X = 3"796. Therefore if the speculator possess 8 florins he may give nearly 3f for the venture. If n = 32, the numerator is '2297, and the denominator is •0570. Hence X = 4'025. Therefore if the speculator have 32 florins he may give rather more than 4 for the venture. If n = 1024 the numerator is "012, and the denominator is '001943. Therefore if the speculator have 1024 florins he may give very little more than 6 florins for the venture. The result at which we have arrived is not to be classed THE OTHER SIDE OP THE BARGAIN. 235 with the arbitrary methods which have been again and again propounded to evade the difficulty of the Petersburg problem and other problems of a similar character. Formulae have often been proposed, which have possessed the one virtue of presenting a finite result in the case of this famous pro- blem, but they have often had no intelligible basis to rest upon, or, if they have been established on sound principles, sufficient care has not been taken to draw a distinguishing line between the significance of the result obtained, and the different result arrived at when the mathematical expec- tation is calculated. We have not assigned any new value to the mathematical expectation ; we have not substituted a new expression for the old ; but we have deduced a separate result, which with- out disturbing the mathematical expectation has a definite meaning of its own. We have found not the fair price at which a contingent prospect may be transferred from one inan to another, but the value which such a prospect has to a man in given circumstances. We have simply deter- mined the terms at which a man may purchase a contingent prospect of advantage, so that by repeating the operation — each time on a scale proportionate to his funds at that time — he may be left neither richer nor poorer when each issue of the venture shall have occurred its own average number of times. By continuing the operation indefinitely, the re- currences of each issue will tend to be proportional to their respective probabilities, and, therefore, the condition we have taken is equivalent to the condition that in the long run the man may expect to be neither richer nor poorer. It would be a great mistake to suppose that the price which one man may prudently give for a venture is the price which the man with whom he is dealing may prudently take 236 CHANCE. for it, or that it is a fair price at which to make the compact. The price which the man may prudently give is not even the price which the same man may prudently take if he change sides with his fellow gambler. The sum in con- sideration of which a man possessed of n pounds may accept a position in which p^ is the chance of his having to pay Pj, p^ the chance of his having to pay p^, amd so on (where Sp = 1) must be obtained by changing the algebraical signs of X, Pj, Pjj.-.in the formulae of Propositions LIX. and LX. Thus on the hypothesis of Proposition LIX. (Cor.) we shall have Z = 2 (pP) + ^ [S ipP') - {S (j>P)n and in the more general case dealt with in Proposition LX. we shall have X n — P^ n — F^ n —P^ CHAPTER X. THE GEOMETRICAL REPRESENTATION OF CHANCES. It is often convenient to represent or depict chances by a geometrical figure. As every chance expressed in terms of certainty is a proper fraction, or a ratio of lesser inequality, it may be represented by the ratio of a smaller length to a larger length, or a smaller area to a larger area, or (in solid geometry) a smaller volume to a larger volume. Especially it is often convenient to represent it as the ratio of a part of a line (or area or volume) to the whole line (or area or volume). And in this case the complementary chance is at the same time represented by the ratio of the remainder to the whole, and the odds are represented by the AB ratio of the part to the remainder. Thus if -jj^ (figure 1) represent the chance of a given event happening, -j^i '^ill be the chance of its not happening, and AB : BG the odds in favour of its happening. Sometimes the question under consideration is geometri- cal in its character, and the result comes out naturally in 238 CHANCE. a geometrical form. For instance, a chain AG (figure 1) Figure 1. A B C consists of a length AB oi iron and a length BG of brass : required the chance that a link taken at random will be iron. The chance is evidently AB -i- AG. So, if a point be taken at random in an area (or volume) a, the chance that it will be within a smaller interior area (or volume) b is expressed by the ratio of the areas, and is 6 -=- a. But in other cases where the question is not of a geometrical nature, the chances involved in it may be con- veniently represented by the ratios of geometrical quan- tities. If an event may happen, or an experiment turn out in several ways not all equally likely, we may represent the probabilities of the several ways by contiguous segments of a straight line, the segments being proportional in length to the respective probabilities. Then if one of the ways must happen the sum of the segments will correspond to certainty, and the ratio of the several segments to the whole will represent the chances of the several ways. PROPOSITION LXII. To express geometrically the probability of an event Z, which depends uptm the way in which a previous event X may have happened, X necessarily happening in one of a finite number of ways. Suppose that X can happen in a variety of ways denoted by A, B, C...M; and let a, b, c.m be the respective chances of these ways ; where a + b + c + ... + 'm=h Also when A has happened let a be the chance of Z GEOMETRICAL REPRESENTATION OF CHANCE. 239 happening ; and when B has happened let /S be the chance of Z; and when G has happened let 7 be the chance of Z, and so on. a b c d m Let the straight line PQ of length unity be divided into segments of the respective lengths a, b, c.m. And on each of these segments describe a rectangle of height unity. And from the rectangles whose bases are a, h, c, &c., cut off segments of height a, /3, 7, &c., and let these segments be shaded. Then the area of the first shaded segment is aa, and it therefore represents the chance that X should happen in the first way (A), and that ^should then happen. So the area of the second shaded segment represents the chance that X should happen in the second way (5), and that Z should then happen. And so on. Hence the whole shaded area (or its ratio to the area of the great square which is unity) will represent the total chance of the event Z, which was required. PROPOSITION LXIII. To express geometrically the prohahility of an event Z, which depends upon the way in which a previous event X shall have happened, X necessarily happening in one {and only one) of an infinitely great number of ways. 240 CHANCE. Make the same construction as in the last case, but with the provision that the number of the segments of PQ is ultimately to be increased indefinitely, and therewith the number of the rectangles upon the segments. If the problem be determinate at all, the height of the shaded rectangles must follow some definite law, and this law will deteimine a curve FGHK such that each rectangle is inscribed with one of its angular points upon the curve. Then (as in Newton, Book I., Lemmas 2 and 3), when the number of the rectangles is increased indefinitely and or their breadth diminished indefinitely, the sum of their areas will be the area bounded by the curved line FGHK, and the straight lines FP, P Q, QK. This curvilinear area (or ' its ratio to the square which is unity) will therefore express the chance required. Example I. A and B each throw once with a common die. What is the chance that A throws higher than B ? Let the straight line AO representing unity be divided into six equal segments AB, BG, CD, BE, EF, FG, repre- senting respectively the (equal) chances of B throwing 1, 2, 3, 4, 5, 6. On each segment erect a rectangle of height unity. Then GEOMETRICAL EEPRESENTATION OF CHANCES. 241 I I I — I — r A B c a E F AB represents the chance of B throwing 1. In this 5 case A's chance of throwing higher is g. Therefore we c may take gths (shaded) of the rectangle on AB to represent A's chance of success in this case. BC represents the chance of B throwing 2. We must 4 take gths (shaded) of the rectangle on BG to represent A'& chance of success in this case. And so on. We thus find A's whole chance of success is represented by the 15 shaded squares in the figure. 15 ^ 36 °^ 12- Note. Of course Es chance is equally ^, and the chance of a tie is g. Example II. The sides of a rectangle are taken at random, each less than 1 inch, and all lengths equally likely. What is the chance that the diagonal is less than 1 inch ? Let HK be a line of length one inch, from which we are to cut off the bsise of the rectangle. Then the ratio of the small segment PQ to the whole Une HK will represent the chance that the end of the base measured from H falls on that segment PQ. w. 16 242 CHANCE. On HK describe a square, and suppose it divided into rectangles on small bases, such as PQ. Let NQ be the rectangle oh PQ. Let a circle from centre H, radius HK, cut PN in B. Then the other side of the rectangle, of which one side is HP, may be any segment of PN, and the diagonal will be greater or less than the radius HR, accord- ing as the segment is greater or less than PR. That is, the diagonal will be less than 1 inch, provided the extremity of the segment of PN lie in PR, and the chance of this is proportional to PR. Hence the question comes under Prop. LXIIL, and the required chance will be the ratio of the quadrant of the circle to the square on HK. That is, the chance is j. Example III. A and B are going to a garden party for an hour each. A will arrive between 3 and 5 p.m., all H P GEOMETEICAL REPRESENTATION OF CHANCES. 243 times within, these limits being equally likely ; and B will arrive between 4 and 7 p.m., aU times equally likely. What is the chance that they are there together ? Take the straight line HK to represent the two hours during which A may arrive, so that the ratio of any segment PQ to the whole HK expresses the chance of A arriving in the interval of time represented by the segment PQ. And on the same scale let HM (at right angles to HK) represent the three hours during which B may arrive. Complete the rectangle MHKN. Let HK be divided into very small segments, to be ultimately increased indefinitely in number and diminished indefinitely in magnitude. And let PQ be one of these segments, and on PQ erect a rectangle QS of height equal to HM. Bisect the right angle MHK by the straight line NRV, cutting PS and KN in R and V. Suppose A arrives in the interval of time denoted by PQ. Say at P. He will leave at the expiration of time HP after 4 p.m. If he is to meet B, B must arrive before the expiration of this time, but B may arrive at any instant in the period HM after 4 p.m. Therefore the chance of their meeting (if A arrives at the time PQ) is HP -=- HM, or PR -i-PS; or it is the ratio of the rectangle QR to the rectangle QS. Hence the series of rectangles on the seg- ments of WK, bounded by the diagonal JVF, will represent by their ratio to the rectangles of height NM, the respective chances of A and B meeting when A arrives at the various instants represented by the segments of HK. Therefore the method of Prop. LXIII. is applicable, and the whole chance of A and B meeting is the ratio of the triangle HKV to the rectangle MK; that is, the chance is g. 16—2 EXEECISES. CHOICE. GROUP I. (on Chap. I.) 1. If there are five routes from London to Cambridge, and three routes from Cambridge to Lincoln, how many ways are there of going from London to Lincoln via Cambridge ? 2. Two hostile companies of 100 men each agreed to settle their dispute by single combat. In how many ways could the two champions be chosen ? 3. Having four seals and five sorts of sealing wax, in how many ways can we seal a letter ? 4. In how many ways can a consonant and a vowel be chosen out of the letters of the word almost? 5. In how many ways can a consonant and a vowel be chosen out of the letters of the word orange? 6. A die of six faces and a teetotum of eight faces are thrown. In how many ways can they fall ? 7. There are five routes to the top of a mountain, in how many ways can a person go up and down ? 8. Out of 20 knives and 24 forks, in how many ways can a man choose a knife and fork? And then, in how many ways can another man take another knife and fork? 9. Out of a list of 12 regular and 5 irregular verbs in how many ways can we choose an example of each? EXERCISES. 245 10. Out of 12 maseuline words, 9 feminine and 10 neuter, in how many ways can we choose an example of each ? 11. Having six pairs of gloves, in how many ways can you take a right- hand and a left-hand glove without taking a pair ? 12. Out of 5 bibles and 12 prayer books, in how many ways can I select a bible and prayer book ? 13. Out of three bibles, seven prayer books and seven hymn books, in how many ways can I select a bible, prayer book and hymn book ? 14. A bookseller shews me six bibles, three prayer books and four hymn books, also five volumes each containing a bible and a prayer book bound together, and seven volumes each containing a prayer book and hynm book bound together. In how many ways can I select a bible, prayer book and hymn book ? 15. If in addition the bookseller produced three volumes, each con- taining a bible and a hymn book, what would my choice now be ? 16. A basket contains 12 apples and 10 oranges, John is to choose an apple or an orange, and then Tom is to choose an apple and an orange. Shew that if John chooses an apple, Tom will have more choice than if John chooses an orange. 17. How many changes can be rung upon eight bells? And in how many of these will an assigned bell be rung last ? 18. There are three teetotums, having respectively 6, 8, 10 sides. In how many ways can they fall? and in how many of these will at least two aces be turned up ? 19. Having bunting of five different colours, in how many ways can I select three colours for a tricolour flag ? 20. In how many ways can I make the tricolour, the three selected colours being arranged in horizontal order ? GEOUP II. (on Chap. I.) 21. A pubhsher proposes to issue a set of dictionaries to translate from any one language to any other. If he confines his system to five languages, how many dictionaries must he publish ? 22. If he extend his system to ten languages, how many more dictionaries must he issue? 23. A man lives within reach of three boys' schools and four girls' schools. In how many ways can he send his three sons and two daughters to school? 246 CHOICE. 24. If there be 300 Christian names, in how many ways can a child be named without giving it more than three Christian names? 25. In how many ways can four persons sit at a round table, so that all shaU not have the same neighbours in any two arrangements 1 26. In how many ways can seven persons sit as in the last question? And in how many of these will two assigned persons be neighbours ? And in how many will an assigned person have the same two neighbours 1 27. Five ladies and three gentlemen are going to play at croquet; in how many ways can they divide themselves into sides of four each, so that the gentlemen may not be all on one side? 28. I have six letters to be delivered, and three boys offer their services to deliver them. In how many ways haye I the choice of sending the letters? 29. A has seven different books, B has nine different books ; in how many ways can one of A'z books be exchanged for one of B's ? 30. In the case of the last question, in how many ways can two books be exchanged for two ? 81. Five men, A,B, G, D, E, are going to speak at a meeting; in how many ways can they take their turns without B speaking before A ? .32. In how many ways, so that A speaks immediately before B ? 33. A company of soldiers consists of three officers, four sergeants and sixty privates. In how many ways can a detachment be made consisting of an officer, two sergeants and twenty privates ? In how many of these ways will the captain and the senior sergeant appear ? 34. Out of a party of twelve ladies and fifteen gentlemen, in how many ways can four gentlemen and four ladies be selected for a dance ? 35. A man belongs to a club of thirty members, and every day he invites five members to dine with him, making a different party every day. For how many days can he do this ? 36. Shew that the letters of anticipation can be arranged in three times as many ways as the letters ot commencem,ent. 37. How many five-lettered words can be made out of 26 letters, repe- titions being allowed, but no consecutive repetitions (i.e. no letter must foUow itself in the same word) ? 38. A boat's crew consists of eight men, of whom two can only row on the stroke side of the boat, and three only on the bow side. In how many ways can the crew be arranged? 39. Having three copies of one book, two copies of a second book, and one copy of a third book, in how many ways can I give them to a class of EXEECISES. 247 twelve boys, (1) so that no boy receives more than one book; and (2) so that no boy receives more than one copy of any book? 40. In how many ways can a set of twelve black and twelve white draught-men be placed on the black squares of a draught-board? GROUP III. (ON Chap. I.) 41. In how many ways can the letters of the word possessions be arranged? 42. In how many ways can the letters of cockatoohe arranged ? 43. Shew that the letters ot cocoon can be arranged in twice as. many ways as the letters oicocoa. 44. In how many ways can the letters oi pallmallhe arranged with- out letting aU the Vb come together? 45. In how many ways can the letters of oiseau be arranged so as to have the vowels in their natural order ? 46. In how many ways can the letters of coco a be arranged so that a may have the middle place? 47. In how many ways can the letters oi quartushe arranged so that q may be followed by «? 48. In how many ways can the letters of ubiquitoushe arranged so that 2 may be followed by m ? 49. In how many ways may the letters of quisquis be ajranged So that each q may be followed by m ? 50. In how many ways can the letters of indivisibility be arranged without letting two i's come together ? 51. In how many ways can the letters of fa eetioushe arranged with- out two vowels coming together? 52. In how many ways can the letters of facetious be rearranged without changing the order of the vowels? 53. In how many ways can the letters of abstemiously Toe rearranged without changing the order of the vowels ? 54. In how many ways can the letters of parallelism be rearranged without changing the order of the vowels ? 55. In how many ways can the letters of almosthe rearranged, keeping the vowels at their present distance apart ? 56. In how many ways can the letters of logarithms be rearranged, so that the second, fourth and sixth places may be occupied by consonants ? 248 CHOICE. 57; In how many ways can two consonants and a vowel be chosen out of the word logarithms, and in how many of these will the letter s occur? 58. In how many ways can the letters of syzygy\>e arranged without letting the three y'e come together ? 59. In how many ways without letting two ^'s come together? 60. In how many ways can we arrange the letters of the words choice and chance without letting two c's come together? GBOUP IV. (OS Chap I.) 61. In how many ways can ten books be made into five parcels of two books each? 62. In how many ways can nine books be made up in four parcels of two books each and one book over ? 63. In how many ways can they be made into three parcels of three books each? 64. Ten men and their wives are to be formed into five parties, each consisting of two men and two women. In how many ways can this be done? 65. In how many of these ways will an assigned man find himself in the same party with his wife ? 66. In how many will two assigned men and their wives be together in one party? 67. In how many ways can a purchaser select half a dozen handkerchiefs at a shop where seven sorts are kept ? 68. A choir contains 10 members. In how many ways can a different six be selected every day for three days ? 69. A man has six friends and he invites three of them to dinner every day for twenty days. In how many ways can he do this without having the same party twice ? 70. Find the total number of selections that can be made out of the letters ned needs nineteen nets, 71. Find the total number of selections that can be made out of the letters daddy did a deadly deed. 72. How many selections of three letters can be made out of the letters in the last question ? 73. How many arrangements of three letters can be made out of the same ? EXERCISES. 249 74. Shew that there are 8 combinations 3-together of the letters veneer, and 16 combinations 4-together of the letters veneered. 75. How many are the combinations 3-together of the letters weddedt 76. How many are the combinations 4-together of the letters re- deemedl 77. The number of combinations 5-together of the letters ever esteemed is 32. 78. In how many ways can three persons divide among themselves the letters ever esteemed? 79. In how many ways can they divide them so that each may take four? 80. A has the letters esteem, B has feeble, and G has veneer. In how many ways can they exchange so that each shall have sis 1 GEOUP V. (on Chap. I.) 81. In how many ways can the letters oi falsity be arranged, keeping the consonants in their natural order, and the vowels also in their natural order ? 82. In how many ways can the letters of affection be arranged, keeping the vowels in their natural order and not letting the two /'s come together ? 83. In how many ways can the letters of kaffeekanne be arranged so that the consonants and vowels come alternately ? 84. In how many ways can the letters of delete be arranged, keeping the consonants in their natural order? 85. In how many ways can the letters of delete be arranged, keeping the consonants in their natural order and not letting two e's come together ? 86. In how many ways can the letters of delirious be arranged, keeping the consonants in their natural order and the vowels in their natural order ? 87. What would be the answer of the last question if the two i's must not come together? 88. In how many ways can the letters of fulfil be arranged without letting two letters which are alike come together ? 89. In how many ways can the letters oi murmur he arranged without letting two letters which are alike come together ? 90. How many selections of four letters can be made out of the letters of murmur; and how many arrangements of four letters? 250 . CHOICE. 91. How many arrangements of four letters can be made oiit of the letters of /Mi/iZ. 2. 92. How many arrangements of five letters can be made out of the letters oipallmall? 93. How many arrangements of four letters can be made out of the letters of k affe e Jeanne without letting the three e's come together ? 94. How many arrangements of six letters can be made out of the letters of nineteen tennis nets? 95. How many arrangements of six letters can be made out of the letters of little pipe without letting two letters which are alike come together ? 96. How many arrangements of five letters can be made out of the letters of murmur er without letting the three r's come together ? 97. In how many ways can the letters of quisquis be arranged without letting two letters which are alike come together ? 98. In how many ways can the .letters ot feminine be arranged without letting two letters which are alike come together ? 99. In how many ways can the letters of muhamm,acLan\>e arranged without letting three letters which are alike come together ? 100. In how many ways can the same letters be arranged without letting two letters which are alike come together ? GEOUP VI. (on Chap. I.) 101. Out of 20 consecutive numbers, in how many ways can two be selected whose sum shall be odd ? 102. Out of 30 consecutive integers, in how many ways can three be selected whose sum shall be even ? 103. Out of an unlimited number of pence, half-pence and farthings, in how many ways can twenty coins be selected ? 104. A person holds five coins and asks you to guess what they are. Knowing that he can only have sovereigns, half-sovereigns, crowns, half- crowns, florins, shillings, sixpences, fourpences, or threepences, how many wrong guesses is it possible to make 1 105. In the ordinary scale of notation how many numbers are there con- sisting of five digits ? In how many of these is every digit an odd number ? In how many is every digit an even number? In how many is there no digit lower than 6 ? In how many is there no digit higher than 3 ? How many contain aU the digits 1, 2, 3,4, 5 ? and how many contain all the digits 0, 2, 4, 6, 8 ? EXERCISES. 251 106. How many different sums can be thrown with two dice, the faces of each die being numbered 0, 1, 3, 7, 15, 31 ? 107. How many different sums can be thrown with three dice, the faces of each die being numbered 1, 4, 13, 40, 121, 364? 108. At a post-office they keep ten sorts of postage-stamps. In how many ways can a person buy twelve stamps? In how many ways can he buy eight stamps? In how many ways can he buy eight different stamps? 109. In how many ways can a pack of 52 cards be dealt to 13 players, four to each, (1) so that every one may have a card of each suit, (2) so that one may have a card of each suit, and no one else have cards of more than one suit ? 110. In how many ways can a pack of cards be dealt to four players, subject to the condition that each player shall have three cards of each of three suits and four cards of the remaining suit ? 111. In how many ways can 18 different things be divided among five persons, (1) so that four of them have four each, and the fifth has two things ; (2) so that three of them have four each and the other two have three each ? 112. If there be 14 sorts of things, and two things of each sort, find the total number of selections that can be made. 113. If there be twenty sorts of things and nine things of each sort, shew that 99999999999999999999 different selections can be made. 114. The game of bagatelle is played with 8 balls aU alike and 1 different. The object is to get as many as possible of these balls into 9 different holes, no hole being capable of more than one ball. How many different dispositions of balls in the holes are possible ? 115. If there were 7 balls alike and 2 others alike, how may dispositions would be possible ? 116. If there were 7 balls alike and 2 others different, how many dispositions would be possible? 117. In how many ways can 27 different books be distributed to A, B, C, so that A and G together may have twice as many as B ? 118. Shew that out of 99 things the number of ways of selecting 70 is to the number of ways of selecting 30 as 3 to 7. 119. Four boys are in attendance at a telegraph office when 8 messages arrive. In how many ways can the messages be given to the boys without leaving any boy unemployed ? 120. Out of the first 100 integers, in how many ways can three be selected whose sum shall be divisible by 3 ? 252 CHOICE. GROUP Vn. (ON Chap. II.) 121. The number of ways of selecting a; things out of 2a; + 2 is to the number of ways of selecting x things out of 2x - 2 as 99 to 7. Find x, 122. If (^+iJ^=P?, findn. If (^+JJ?=|P?, find re. 123. IfBi!=mi^and(7Ei3'=reC^, findaandy. 124. Shew that CJ+' : Br^=(^-' : iJS+^ 125. Shew that {n-r)R^ a; = nK C>'. 126. Shewthat {c"4;-c;}C!;:}-7-{(C",)^-c?'^}c;:l}=?-. 127. Shew that out of m+ra- 1 things the choice in selecting m things is to the choice in selecting n things as n to m. 128. There are m parcels, of which the first contains n things; the second 2m things ; the third Sn things ; and so on. Shew that the number of ways of taking n things out of each parcel is Imre -=■ {Ire}™. 129. Shew that the number of ways of dividing 2re+2 things into two equal parcels is the sum of the number of combinations of 2re things, re and re - 1 together. 130. Shew that when repetitions are allowed the number of selections of re things out of m+ 1 is the same as the number of selections of m things out of re+1. 131. Shew that JB^=2B^ and iJf=3i?5„. 132. When repetitions are allowed the choice of m things out of n is to the choice of re tilings out of m as re to m. 133. A basket contains 2re + r apples and 2n-r pears. Shew that the choice of re apples and re pears is greatest when r=0, n being constant. 134 Out of 3re consecutive integers, in how.many ways can three be se- lected whose sum shall be divisible by 3 ? 135. If pq + r different things are to be divided as equally as possible among J) persons, in how many ways can it be done? {r<:p). 136. Prove that the number of ways in which p positive signs and n negative signs may be placed in a row, so that no two negative signs shall be together, is equal to the number of combinations of ^i + 1 things taken re together. {p>n). 137. The number of boys in the several classes of a school are in arith- metical progression, and a number of prizes equal to the common difference of the progression is to be given to each class, no boy receiving more than EXERCISES. 253 one prize. Shew that if the prizes are all different the number of ways of giving them is the same as if all were to be given to the largest class. 138. The number of different throws that can be made with n dice is 139. The number of ways of selecting four things out of n different things is one-sixth of the number of ways of selecting four things out of 2n things which are two and two alike of n sorts : find n. 140. If bagatelle is played with n balls alike and one different, and there are n+1 holes each capable of receiving one ball, the whole number of ways in which the balls can be disposed is (n+ 3) 2" - 1. GEOUP vni. 141. In how many ways can a triangle be formed, having its angular points at three of the angular points of a given hexagon ? 142. How many triangles can be formed having every side either 4 or 5 or 6 or 7 inches long ? 143. How many different rectangular parallelepipeds can be constructed, the length of each edge being an integral number of inches not exceed- ing 10? 144. If four straight lines be drawn in a plane and produced indefinitely, how many points of intersection will there generally be ? 145. If four straight lines be drawn in a plane, no two being parallel and no three concurrent, how many triangles will they form ? 146. There are n points in a plane, no three being in a straight line, except p of them which are all in a straight line ; how many triangles can be formed having some of these points for vertices ? 147. There are p points in a straight Une, and g points on a parallel straight line ; how many triangles can be formed having some of these points for vertices ? 148. If there be r more points lying on another parallel straight line, how many additional triangles can be formed, assuming that no three points lie on any transverse straight line ? 149. Each side of a square is divided into n parts. How many triangles can be formed having their vertices at points of section ? 150. If n straight lines be drawn in a plane, no two being parallel and no three concurrent, how many points of intersection will there be ? 254 CHOICE. 151. If n straight lines be drawn in a plane, no two being parallel and no three concurrent, except p, which meet in one point, and q which meet in another point, how many other points of intersection will there be ? 152. Into how many parts is an infinite plane divided by n straight lines, of which no three are concurrent and no two parallel ? 153. Into how many parts is infinite space divided by n planes, of which no four meet in a point and no two are parallel ? 154. A large floor is paved with sguare and triangular tiles, all the sides of the squares and triangles being equal. Every square is adjacent to four triangles, and every triangle is adjacent to two squares and a triangle. Shew that the number of angular points must be the same as the number of triangles, and that this must be double of the number of squares. 155. If, in a different pattern, the triangles be placed in blocks of three forming a trapezium, and every trapezium be adjacent to five squares, the same results will hold good as in the last question. 156. In how many ways can we form a triangle having each of its sides an integral number of inches, greater than n and not greater than in ? 157. How many of these wiU be isosceles, and how many equilateral ? 158. The number of triangles that can be formed having each of their sides an integral number of inches not exceeding 2ra is ^n(n+V) (4n+5). And excluding all equilateral and isosceles triangles it is i w (n - 1) (4™ - 5). 159. If the number of inches is not to exceed 2m - 1 the number of triangles will be ^re(re + l) (4n- 1). And excluding all equilateral and isosceles triangles it will be ^ (» - 1) (n - 2) (4ra - 3). 160. If there be n straight lines in one plane, no three of which meet in a point, the number of groups of n of their points of intersection, in each of which no three points lie in one of the straight lines, is i In - 1. GROUP IX. 161. The number of ways of dividing 2n different things into two equal parts, is to the number of ways of similarly dividing 4n different things, as the continued product of the first n odd numbers to the continued product of the n odd numbers succeeding. 162. Shew that there are 2"-i ways of selecting an odd number of things out of n things. 163. If there be one card marked 1, two marked 2, three marked 8, and so on : the number of ways of selecting two cards on which the sum of the EXERCISES. 255 numbers shall be n is f^n (n^ - 1) or jV* (m" - 4) aeoording as n is odd or even. 164. There are 3m + 1 things of which n are alike and the rest all differ- ent : shew that there are 2^" ways of selecting n things out of them. 165. In how many ways can three numbers in arithmetical progression be selected from the series 1, 2, 3...2n, and in how many ways from the series 1, 2, 3...(2n+l)? 166. By considering the selections of n things out of n things with repeti- tions, writing down separately the number of selections when 1, 2, 3... or m different things are selected, shew that 167. Find the number of ways in which 37i different books may be distributed to three different persons so that their shares may be in Arithmetical Progression, ..(The cases in which the shares would be 0, «, 2re or n, n, n are to be included.) 168. If M^ be the number of permutations of m things taken r together, and N^ the number of permutations of n things taken r together, prove that the number of permutations of m + n things r together will be ob- tained by expanding {M+N)'', and in the result replacing the indices by suffixes. 169. In the expansion of ((Sj-l-ffij-l- ... -i- Oj,)" where n is an integer not greater than p, there are CJ terms, in none of which any one of the quantities a^, a^-.-Op occurs more than once as a factor; and the coefBcient of each of these terms is In. 170. There are 2n letters, two and two alike of n different sorts. Shew that the number of orders in which they may be arranged, so that no two letters which are alike may come together, is -., |.,.-^'2|2n-l-h'^g^^2^ |2«-2 -&c. to n-i-1 termsl. 2»|l — 1 171. The number of ways in which r different things may be distributed among n+p persons so that certain n of those persons may have one at least is {n+pY -n{n+p-lf+''^^^ (n+p~2)'- -&a. 172. Shew that, for n different things, 1 - (number of partitions into two parts)-!- 12 (number of partitions into three parts) - ... ± Ib- 1 (number of partitions into n parts) =0. 256 CHOICE. 173. Shew that, if m>n, SS{i^-=-i?} = (m + l)-r-(m-n + l). 174. Shew that SS {c:C-i- C^}= (2n + 1)-H(n+ 1).' 175. Prove that, to n terms, and iJi" + a'iff + 3=JC + . . . = (n + l)''iC+' - (2m + 3) JJr*^ + 2iC^- Verify the results when s=2. 176. Prove that m m(m+l) m(m + iy(m + 2) . , = ^ + ^i|L+l) H- "("+l)("^+2) + ... to ^ terms. . 177. Shew that |2re-l -f- ( |m - 1)= is the sum to n terms of the series (C5)2 + 2(CS)» + 3(C?)»+.... 178. Apply Prop. xiv. to shew that \n+r-l „\n+r-3 mfe-H |m+r-5 \n\n-l I -^ 1 + ''<"/JI -&c. (tiU it Stops) = H ■ |r 1 |r-2 1.2 [r-4 ■ ^ ' |r | ra-r Or, putting ra=r, |2«-1 „ 1^-3 „(„_!) J2«-5 ^ Ib 1 |m-2 "^ 1.2 |re-4 I : 179. Shew that, to n terms. \n + a \a \ a + l \a+2 ^ \n+b-l \b-l )6 "^ ]6 + l "^ ]6 + 2 '^•••^ a-6+1 180. Shew that, to n terms, la + l |t+-i)_ and the number of terms being the integer next less than {mn + m + Mj-i-{2n + 1). 199. In how many ways can five black balls, five red balls, and five white balls be distributed into three different bags, five into each? 200. If there be 3 sorts of things, and re things of each sort, the number of ways in which three persons can divide them so as to have n things each is C^-K Cj"+?-3 Cj+s = -I (n + 1) (n + 2) (n^ + Sn+i). GROUP XI. 201. In how many ways can six Englishmen, seven Russians, and ten Turks be arranged in a row, so that each Englishman may stand between a Russian and a Turk, and no Russian and Turk may stand together? 202. In how many ways can five Englishmen, seven Russians, and ten Turks be arranged in a row, so that each Englishman may stand between a Russian and a Turk, and no Russian and Turk may stand together? 203. Find the number of combinations that can be formed out of the letters of the following Une (Soph. PMloct. 746) : airainrairac TairairiraTraTnTairainTairaiy taking them (1) 5 together, and (2) 25 together. 204. "What is the total number of combinations that can be made out of the same letters ? 205. How many solutions can be given to the following problem? "Find two numbers whose greatest common measure shall be G and their least common multiple M= Ga°- b^ c^ d' ; u,, 6, c, d being prime numbers." 206. How many solutions can be given to the following problem? "Find two numbers of which G shall be a common measure, and M (as in the last question) a common multiple." EXERCISES. 259 207. Out of 20 letters how many combinations of six letters can be made, no letter occurring more than twice in the same combination? 208. A certain symmetrical function of x, y, z is such that x is found in five of its terms, xy in three terms, xyz in two terms, and there is one term without x, or y, or 2. How many terms are there altogether ? 209. I have six friends, each of whom I have met at dinner 12 times. 1 have met every two of them 6 times, every three of them 4 times, and every four of them 3 times, every five twice, all six once, and I have dined out eight times without meeting any of them. How many times have I dined out altogether ? 210. Two examiners, working simultaneously, examine a class of 12 boys, the one in classics the other in mathematics. The boys are examined individually for five minutes each in each subject. In how many ways can a suitable arrangement be made so that no boy may be wanted by both examiners at once? 211. Out of six pairs of gloves, in how many ways can six persons take «ach a right-handed and a left-handed glove without any person taking a pair? 212. Out of nine pairs of gloves, in how many ways can six persons take €ach a right-handed and a left-handed glove without any person taking a pair? 213. A square is divided into 16 equal squares by vertical and horizontal lines. In how many ways can 4 of these be painted white, 4 black, 4 red, and 4 blue, without repeating the same colour in the same vertical or hori- zontal row ? 214. The number of ways of deranging a row oi m+n terms so that m are displaced and n not displaced is \m+n llm Im In 215. Shew that, if n be an integer, Ib^ is divisible by ( [n)"+' ; and if m and n are odd numbers \mn is divisible by (|m) ^ (|«) = . 216. If there be an odd number of dominoes in a set that begins with double-blank and if the total number of pips on the whole set be even, the number of pips on the highest domino will be a multiple of 8. 217. If /„ denote the number of derangements of n terms in circular j)roeession so that no term may follow the term which it followed originally, /»+/n+l= Wlh 17—2 260 CHOICE. 218. Find the number of positive integral solutions of the equation x+y+ii+ ... {p variables) =m, the variables being restricted to lie between 2 and n, both inclusive. 219. If there be seven copies of one book, eight of another, and nine of another, in how many ways can two persons divide them, each taking 12 books? 220. The squares on a chess-board are painted with 8 colours, 8 squares with each colour. The colours are to be arranged so that every horizontal line contains every colour, but the vertical lines are only subject to the condition that no two adjacent squares must be alike. How many arrange- ments are possible ? GROUP XII. 221. There are n things alike and n others all different; shew that there are 2" ways of selecting n things out of them, and that the number of orders in which all the 2m things can be arranged is 2" . 1 . 3 . 5. ..(2m- 1). 222. If a set of dominoes be made from double blank up to double n, prove that the number of them whose pips are m-r is the same as the num- ber whose pips are n + r, and this number is i(2m-2j'-|-3±l) ; and the total number of dominoes is ^ (m-l-1) (m-H2). 223. If all the combinations of n letters (1, 2, 3. ..or n together) be written down, the number of times each letter wUl occur is 2'^'. 224. The total number of permutations [1, 2, 3. ..or {m+n) together} which can be made out of m things all alike and n other things all alike is (with the notation of page 70) Gf^i^^^i - 2. 225. If we write down all possible permutations [1, 2, 3... or (m+ji) together] of m a's and m ;8's, we shall write a and we shall write /3 , mm-l-m-1 ^ — m+2 — ^'»+i."+i times; , 7mi + n- 1 „ "'' m + 2 "^^ "+' ^^• 226. Write down all the permutations 1, 2, 3, 4 or 5 together of the letters mamma, and verify the result of the last question. 227. The total number of permutations [1, 2, 3. ..or {m+n + 1) together] which can be made out of m things alike, n other things alike, and one different thing, always using the one different thing is, , mn+m + n EXERCISES. 261 228. The total number of permutations [1, 2, 3 or...(m+w+l) together] ■which can be made out of m things alike, n other things alike, and one different thing, is (m+l)(«+l) m + n + 3 ^'-*^'^-^- 229. Write down all the permutations, one or more together, of the letters peppe r, and verify the result of the last question. 230. The total number of permutations of n things [1, 2, 3. ..or n "together] is the integer nearest to c In - 1. 231. The total number of permutations of n things [1, 2, 3,. ..or n together] is greater than e I » - 1 — and less than e I m - 1 =- . 232. If all the permutations of n letters (1, 2, 3. ..or n together) be ■written do^wn, the number of times each letter ■will occur is the integer nearest to « (m - 1) I ra - 1. 233. The integer in the last question is always greater than e (» - 1) |» - 1 2 -and less than e(re-l)|n-lH — 5 — -. 234. In how many different orders can seven men and five women get into a carriage, one by one, so that there may never be more women than men in it ? 235. In how many orders can m positive units and 11, negative units be arranged so that the sum to any number of terms may never be negative? 236. In how many orders can m even powers of x and n odd powers of x be arranged so that when x=-l the sum to any number of terms may never be negative ? 237. In how many orders can a man ■win m games and lose n games so as at no period to have lost more than he has won ? 238. In how many different orders can a man possessed of h pounds •win m wagers and lose n wagers of 1 pound each without being ruined during the process ? 239. If p, q, r be integers, such that the two lowest are together greater than the highest, and p + q+r='is, the number of ways in which -two men can divide p black balls, q white and r red, so that each may take s baUs, is 8^ + s + 1 - J (;)2 + j2 + ^2). 240. If the two lowest are together less than the highest (q+r6, A'a chance is 1 — - — , B'a chance is -= — , and the chance of a tie is - . 2a 2a a 312. If in the last exercise they are to spin again in case of a tie, A'& chance of winning will be — = =— , and B'a chance will be ^ = . 2a — 2 2a - 2 313. Out of 2n tickets, numbered consecutively, three are drawn at random. Find the chance that the numbers on them are in a. p. 314. Out of 2» + 1 tickets, consecutively numbered, three are drawn at random. Find the chance that the numbers on them are in a. p. 315. A and B shoot at a mark, and A hits it once in n times, and B once in n - 1 times. If they shoot alternately, A commencing, compare their chances of first hitting the mark. 316. If there be m sorts of things and n things of each sort, shew that the chance that m-r things selected at random may be all different is 317. If « things (a, /3, y, &o.) be arranged in a row, subject to the con- dition that a comes before /3, what is the chance that a comes next before /S ? 318. There are « counters marked with odd numbers, and n more marked with even numbers ; if two are drawn at random shew that the odds are n to « - 1 against the sum of the numbers drawn being even. 319. If a head counts for one and a tail for two, shew that Sn is the most likely number to throw when 2n coins are tossed. Also shew that the chance of throwing 3 (n+1) with 2(b+1) coins is less than the chance of throwing 3m with 27i coins in the ratio 2b +1 . 2n + 2. 320. A red card has been removed from a pack. Thirteen cards are then drawn and found to be all of one colour. Shew that the odds are 2 to 1 that this colour is black. 268 CHANCE. GEOUP XVII. 321. Two squares are marked at random on a chess board. Shew that the odds are 17 to 1 against their being adjacent. 322. If two squares of opposite colour are marked at random on a chess board the chance that they are adjacent is jt ? 328. If two squares are marked at random on a chess board the chance that they have. contact at a corner is 7 ? 324. If two squares of the same colour are marked at random on a chess board the chance that they have contact at a comer is J^ ? 325. Two squares are chosen at random on a chess-board : what is the chance that one is a castle's move from the other ? 826. What is the chance that both are the same colour, and one a ■castle's move from the other ? 327. Two squares are chosen at random on a chess-board : what is the chance that one is a knight's move from the other ? 328. What is the chance that one is a bishop's move from the other ? 329. What is the chance that one is a queen's move from the other ? 330. What is the chance that one is a king's move from the other ? 331. Pour squares are chosen at random on a chess-board : what is the •chance that they should form a knight's circuit ? (N.B. A knight's circuit is a set of four squares such that a knight can move from one to another, ending where he began.) 832. Two black squares and two white squares are chosen at random on a chess-board : what is the chance that they should form a knight's circuit ? 333. Two boards are divided into squares coloured alternately Uke a chess-board. One has 9 rows of 16 squares each. The other has 12 rows of 12 squares. If two squares are chosen at random, shew that they are more likely to be adjacent on the square board than on the oblong one, in the ratio of 264 to 268 ? 334. In the last question if the squares were chosen at random of opposite colours, the square board would still have the same advantage over the oblong ? 335. The two squares in Question 333 are more likely to have comer- contact on the square board than on the oblong one, in the ratio of 121 to 120? EXERCISES. 269 336. If three squares be chosen at random on a chess-board, the chance that they are in the same line, excluding diagonal lines, is ^j ? 337. The chance that they should be in a diagonal line, is -yxj ' 338. If four squares be chosen at random on a chess-board, find the chance that they be at the corners of a square having its sides parallel to the sides of the board. 339. Find the chance that they be at the corners of a square having its sides parallel to the diagonals of the board. 340. What would be the chance in the last question if the four squares were taken at random of one colour ? GEOUP xvni. 341. In a single throw with four dice, what is the chance of throwing two doublets ? 342. What is the chance of the throw containing a single doublet ? 343. If six dice are thrown together, what is the chance that the throw will be (1) three doublets, (2) two triplets, (3) a quartett and a doublet ? 344. Compare the chances of throwing four with one die, eight with two dice, and twelve with three dice, having two trials in each case. 345. What is the probability of throwing not more than eight in a single throw with three dice ? 346. There is a greater chance of throwing nine in a single throw with three dice than with two dice. Shew that the chances are as 25 to 24. 347. In a single throw with three dice, the chance that the sum turned up will be a multiple of three is one-third ; and the chance that it will be a multiple of six is one-sixth. 348. A prize is to be won by A as soon as he throws five with two dice, or by jB as soon as he throws ten with three dice. If they throw alternately, A commencing, shew that their chances of winning are equal. 349. A and B have each a pair of dice : shew that the odds are nearly 19 to 1 against their making the same throw in one trial. And if they throw with three dice each, the odds are nearly 46 to 1. 350. The chance of throwing 14 is greater with three dice than with five as 54 to 49. 351. The chance of throwing 15 is less with three dice than with five as 36 to 55. 352. If we throw with two dice and count the difference of the two 270 CHANCE. numbers which turn up, this is as likely as not to be one-or-two. Also the chance of zero is the same as the chance of three, and is the same as the chance of four-or-five. 353. A, B, C, D throw with two dice in succession, the highest throw winning a prize of one guinea. If two or more throw equal sums (higher than their competitors) they divide the prize. A having thrown 9, what is the value of his expectation ? 354. A die is loaded so that six turns up twice as often as ace and three times as often as any other face. Another die is loaded so that six turns up three times as often as ace and twice as often as any other face. If these •dice be thrown together what is the chance of throwing sixes ? 355. What is the chance of throwing at least one six ? 356. What is the chance of throwing at least eleven ? 357. What is the chance of throwing doublets ? 358. What is the chance of throwing seven ? 859. Shew that if each die is loaded so that six turns up oftener than any other face in the ratio 7 : 4, the odds are 41 to 40 in favour of a throw with two dice exceeding seven ? 860. If each be so loaded that six turns up less often than any other face in the ratio 2 : 5 the odds wiU be 55 to 26 against the throw exceed- ing seven. GBOUP XIX. 361. Eleven cards are marked with the letters of the word Hammer- smith, and one is lost. ■ Out of the remaining ten a card is drawn and is found to be m. What is the chance that the lost letter was m? 362. If in the last question the drawn card is replaced and again a card drawn at random is found to be m, what is the chance that the lost letter was m? And if the same operation be repeated again with the same result, what does the chance then become? 363. If in the last question the drawn card had not been replaced, what would the chances have been after each operation? 364. It is not known whether a set of dominoes goes to double eight or double nine, each being equally likely. Two dominoes are drawn and found to contain no number higher than eight. Shew that the odds are now three to two in favour of the set going only to double eight. 365. Two letters have fallen out of the word Mis sis s ipp i ; they are picked up and placed at random in the two blank spaces. Find the chance that the letters are in right order. EXERCISES. 271 366. A box contains four dice two of which are true, and the others are so loaded that with either of them the chance of throwing six is i and the chance of throwing ace is -jV . I take two dice at random out of the box and throw them. If they turn up sixes find the chance (1) that both are loaded, (2) that one only is loaded, and (3) that neither is loaded. 367. What would be the chances in the last question, if six and Jive had turned up ? 368. "What would be the chances in the same question if six and one had turned up? 369. One card out of a pack has been lost. From the remainder of the pack thirteen cards are drawn at random, and are found to consist of two spades, three clubs, four hearts, and four diamonds. What are the respect- ive chances that the missing card is a spade, a club, a heart, or a diamond? 370. A man has left his umbrella in one of three shops which he visited in succession. He is in the habit of leaving it, on an average, once in every four times that he goes to a shop. Find the chance of his having left it in the first, second, and third shops respectively. 371. Eleven cricketers had to elect a captain. Bach was as likely as not to vote for himself. Otherwise he voted at random. What are the odds that a man who polled five votes, voted for himself? 372. The odds are estimated as 2 to 1 that a man wiU write rigorous rather than rigourous. Out of the word which he writes a letter is taken at random and is found to be u. What are now the odds that he wrote the word in the former way? 373. If two letters had been taken at random out of the word and found to be both alike, what would be the odds that the word was written in the former way? 374. The face of a die, which should have been marked ace, has been accidentally marked with one of the other five numbers. A six is thrown twice in two throws. What is the chance that the third throw will give a six ? 375. Eeference is made to a month which contains portions of six differ- ent weeks : what is the chance that it contains thirty-one days? 376. One card has been lost out of a pack. From the remainder 13 cards are drawn and are found to be all black. Find the chance that the lost card is red. 377. A bag contains seven cards inscribed with the letters of the word singing. Another bag contains the letters of morning, and a third bag the letters of evening. From one of the bags two letters are drawn and are found to be n and g. What is the chance that it was the first bag? 272 CHANCE. 378. From one of the bags in the last question two letters are drawn and are both alike, what are the respective chances of the bag having been the first, the second or the third? 379. One letter each ia drawn from two of the bags, and the two letters drawn are found to be both alike. What are the respective chances that the bag not drawn from was the first, the second or the third? 380. In six trials, on an average, A hits the centre 3 times, the outer twice, and misses once. B misses 8 times, hits the outer twice, and the centre once. A man who is equally likely to be 4 or B is observed to fire three shots, two of which hit the centre and one misses. Shew that the odds are now 3 to 1 that the man is A. GBOUP XX. 381. If two letters are taken at random out oi little lilies shew that the odds are 5 to 1 against their being both alike. 382. If three of the letters are taken at random the odds are 6 to 5 against two (at least) being alike. 383. Pour persons draw each a card from an ordinary pack. Find the chance (i) that one card is of each suit : (ii) that no two cards are of equal value : (iii) that one card is of each suit and no two of equal value. 384. Each of four persons draws a card from an ordinary pack. Find the chance that one card is of each suit, and that in addition, on a second drawing, each person shall draw a card of the same suit as before. 885. All the cards marked 1, 2, 3, 4, 5 are dealt into one heap, and all marked 6, 7, 8, 9, 10 into another. If a card be taken at random out of each heap, what is the chance that the sum of the numbers on them is eleven? 886. If two cards be taken out of each heap what is the chance that the sum is twenty- two ? 387. A has a bag of 6 tickets numbered from 2 to 7, JS has a box of dominoes numbered from double 1 to double 6, and C has a pair of common dice. A draws a ticket from his bag, then B draws a domino from his box, then G throws his dice, and they continue to do this in order, without re- placing ticket or domino, until someone has drawn or thrown 7, and he is declared the winner. Compare their respective chances of winning. 388. A man undertakes to win two games consecutively out of three games, playing alternately with two opponents of unequal skill. Shew that it is to his advantage to begin by playing with the more skilful opponent. 889. When A and B play at chess A wins on an average two games out of three. What are the respective chances of A winning each of the following EXERCISES. 273 matches, A undertaking (1) to win two games out of the first three, (2) to win two consecutive games out of the first four, (3) to win four games out of the first six? 390. Two men engage in a match to win two games out of three, the odds in every game being three to two in favour of the player who begins. If the winner of any game always begins the next game, shew that the odds on the match are 69 : 56 in favour of the player who begins the first game. 391. A and B play at chess. If A has the first move the odds are eleven to six in favour of his winning the game, but if B has the first move the odds are only nine to five in A'a favour. It is known that A has won a game, what are the odds that he had first move? 392. In the last Question, if A has the first move in the first game, and the loser of each game plays first in the next game, what is A'b chance' of winning at least two out of the first three games? 393. A marksman's score is made up of centres counting 2 each and outers counting 1 each. He finds that he hits the centre once in five trials and he hits the outer ring as often as he misses altogether. What score may he expect to make in 5 shots, and what is the chance that he makes exactly this? 394. What is the chance that he makes more than the expected score, and what is the chance that he makes less? 395. There are ten counters in a bag marked with numbers. A person is allowed to draw two of them. If the sum of the numbers drawn is an odd number, he receives that number of shillings ; if it is an even number, he pays that number of shillings. Is the value of his expectation greater when the counters are numbered from to 9 or from 1 to 10? 396. A bag contains six shillings and two sovereigns. What is the value of one's expectation if one is allowed to draw till one draws a sovereign? 397. One of two bags contains ten sovereigns, and the other ten shil- lings. One coin is taken out of each and placed in the other. This is re- peated ten times. What is now the expectation of each bag? 398. Four whist players cut for partners (i.e. each draws a card, the two highest to play together and the two lowest together), what is the chance that they will have to cut again? 399. The whole number of possible ways in which four whist players can have 13 cards each is 53644737765488792839237440000. Hence shew that if a party play a hundred millions of games the odds are more than a biUiou to one against their ever having the same distribution of cards twice. w. 18 274 CHANCE. 400. A and B play at cards. A'e skill : B'a :: 3 : 2. Nothing is known as to whether either has a confederate; if one has and the other has not, the odds on the dishonest player are doubled ; B wins 3 games successively, what is the chance he will win the next game? GEOUP SXI. The table of Mortality gwen on page 168 is to he used in tlie exercises of (his grawp. 401. "What is the chance that two persons aged 33 and 57 should both be alive two yeais hence? 402. What is the chance that they should both be dead? 403. Find the chance that one and only one is dead. 404. If it be known that one and only one is aJive, find the chance that the youngest is alive. 405. If at the age of 89 the expectation of life is 2-95 years what is the expectation at the age of 88? 406. What is the expectation at the age of 87? 407. Given that at the age of four the expectation of life is 52"1 shew that this is greater than the expectation either at three or five. 408. I promise to pay a boy's school fees for the next five years provided we both live. The fees are £50 per annum payable yearly in advance. If my age is 52 and the boy's age 13 what is the present value of the boy's expectation? (Interest at 5 per cent.) 409. Twelve months later we are both aUve, for what immediate sum might I fairly compound the future payments? 410. On the hypothesis of p. 169 what would be the expectation at the age of 90 ? 411. Compare the chance of a person aged 20 living to be 60, with the chance of a person aged 60 living to be 80. 412. A person is now aged 20, shew that the chance of his dying before he is 60 is nearly equal to the chance of his dying between 60 and 80. 418. Shew that the odds are 2 to 1 against a child 5 years old living to be 70. 414. Shew that the odds are 2 to 1 (nearly) in favour of a child 5 years old living to be 50. 415. A man emigrated leaving two brothers aged 51 and 61. Five years afterwards he hears that one is aUve and one is dead. Find the odds in favour of the younger one being aUve. EXEECISES. 275 416. Three men are known to have been alive four years ago when their ages were 34, 59 and 70. Find the chance that they are all alive now. 417. Fiad the chance that two and only two are alive now. 418. Find the chance that one and only one is alive now. 419. If it be known that one and only one is alive find the chance that it is the youngest. 420. Find the chance that it is the eldest. GROUP XXII. (CAiajs: seqdenges amk paibs.) Throughout this group a "kamd" Tneans a collection of cards dealt at random out of an ordinary pack qf 62 cards. A "pair" means Inoo cards alike of different suits, as two aces or two hinge. Three ca/rds alike would give three pairs, as they can be taken two and two in three ways. So, four cards alike give six pairs'. A "sequence'* means a set of ca/rds numbered successively, whether qf the sa/me suit or not. In reckoning sequences the kna/iie, queen and king, are regarded as they were 'marked eleven, twelve, wad thirteen. 4S,\. What is the chance that a hand of 5 cards contains at least two aces ? 422. What is the chance that it contains exactly three aces ? 423. What is the chance that it contains a pair ? 424. What is the chance that it contains two pairs and no more ? 425. What is the chance that it contains exactly three pairs ? 426. What is the chance of having six pairs exactly in a hand of six cards ? 427. What is the chance of having seven pairs in a hand of six cards ? 428. What is the chance of having a sequence of three cards and no more in a hand of five cards ? 429. What is the chance that a hand of four cards should be a sequence ? 430. A has a hand of four cards containing three aces. B has a hand of four cards out of the same pack. What is the chance that he has at least three cards aUke ? 431. Two aces, three kings and a queen having been removed from a pack, what is the chance that a hand of four cards should contain at least two alike ? 432. All the sevens having been removed from a pack what is , the chance that a hand of six cards should be a sequence? 18—2 276 CHANCE. 433. What would the chance be, if all the tens were removed instead of all the sevens ? 434. Three cards out of a hand of six are seen to be seven, eight, nine. What is the chance that the whole hand forms a sequence ? 435. What is the chance that it consists of three pairs? 436. What is the chance that it contains no pair? 437. If the three cards seen were seven, eight, ten, what would be the chance of the whole hand forming a sequence ? 438. A and B have each a hand of five cards out of the same pack. If A' a cards form a sequence of five beginning from ace, what is the chance that JB's cards also form a sequence ? 439. A draws a black and a red card from a pack and B draws two black. Shew that A'i chance of drawing a pair is greater than B's in the ratio of 25 : 13. 440. Shew that in the last question B's chance is the same whether ^'s cards be replaced or not. GEOUP XXni. (Loaded Dice.) 441. Shew that the chance of throwing doublets with two dice oneof which is true and the other loaded is the same as if both were true. 442. Shew that the chance of throwing seven is the same as if both were true. 443. There are three dice A, B, C, two of which are true and one is loaded so that in twelve throws it turns up six 8 times, ace once, and each of the other faces twice. Each of the dice is thrown three times and A turns up 6, 6, 1 ; B turns up 6, 5, 4 ; and C turns up 8, 2, 1. What are now the respective chances that A, B ox G is loaded? 444. If the dice in the last question are thrown onee more and A turns up 6, B 1, and G 4, what are now the respective chances that A,B or C is loaded ? 445. If two dice are loaded alike in any way whatever, shew that the chance of throwing doublets is increased by the loading. 446. Shew that the chance of throwing doublets-or-seven is increased by the loading, except in one particular case in which it is unaltered. EXEECISES. 277 447. In the case where two dice are loaded alike, ao that the chance of throwing doublets-or-seven is unaltered, if A+/ be the chance of throwing doublets, the chance of throwing triplets with three such dice will be 448. The chance of throwing three such dice so that two may turn up aUke and the other different is ^+^. 449. The chance of throwing the same three dice so that all may turn up different is -g - 2/. 450. There are 1 + 2" dice of which one is loaded so that the chance of throwing six with it is X ; the rest are all true. If one of the dice be thrown n times and always turns up six, shew that it is as likely as not to be the loaded one. 451. If in the n throws there be only )• sixes the odds will be 5"" to 2""' against the die being the loaded one. 452. If there be two dice, loaded so that with either the chances of throwing 1, 2, 3, 4, 5, 6 are as 1-a; : l + 2a; : 1-x : 1+x : l-2x -. 1 + x, shew that the chances of throwing more-than-seven, seven, and less-than-seven areas5+'^a;2 : 2-iii? : 5 + 2x2. 453. Shew that with three dice loaded as in the last question the odds against throwing more-than-ten are as 18 + a; + Sx' to 18 - a; - 5a^. 454. Shew that the most likely sum to be thrown with 2n dice is 7n: and with 2n+l dice the most likely sums are 2«+3 and 271+4. 455. If n dice be thrown the chance of an even number of aces turning up is one-half of 1 + (#)", and the chance that the number of aces is a multi- ple of 4 is one-fourth of the sum of the nth powers of the roots of the equa- tion (Ba- 5)*= 1. 456. A and B play for a stake which is to be won by hiin who makes the highest score in fonr throws of a die. After two throws, A has scored 12, and B 9. What is ^'s chance of winning? 457. A and B throw for a stake, A having a die whose faces are numbered 10, 13, 16, 20, 21, 25 ; and B a die whose faces are numbered 5, 10, 15, 20, 25, 30. The highest throw to win, and equal throws to go for nothing. Prove that the odds are 17 to 16 in favour of A. ' 458. A die of any number of faces has /i of its faces marked 1, and v marked - 1, the rest being zero. It is repeatedly thrown tmtil the sum of the ntimbers turned up no longer lies between the limits m and -n. Shew that either extreme of these limits is equally likely to be reached if 278 CHANCE. 459. A and B throw each with a pair of dice until one of them throws sixes. A'b dice are true, but one of B's is loaded so that the chance of its turning up six is ^. Shew that if A has the first throw their expectations are equal. 460. A and B throw each with a pair of dice until one of them has thrown six-and-one. A's dice are true, but one of B's is loaded so that six turns up three times as often as any other face. Shew that the odds are 17 to 12 against A if he begins, and 18 to 11 against 4 if B begins. GEOUP XXIV. 461. A bag contains 2ra counters, of which half are marked with odd numbers and half with even numbers, the sum of all the numbers being 5, A man is to draw two counters. If the sum of the numbers drawn be an odd number, he is to receive that number of shillings ; if an even number, he is to pay that number of shillings. Shew that his expectation is worth (in shillings) 8 n{2n-l)' ^ 462. If in the case of the last question there hem+n counters, of which m are marked with odd numbers, amounting to M, and n with even num- bers amounting to N, the man's expectation is worth M+N-{m-n)(M-N) ^(m, + n){m+n-l) 463. A says that B says that C says that D says that a certain coin turned up head. If the odds are a: bin favour of any person making a true report shew that the odds are a^ + Ga'i^+b* to 4a'6+4a6' in favour of the coin having turned up head. 464. In the last question if the evidence came through a chain of n witnesses, shew that the odds in favour of the event would be (0+5)" + (a -6)" to (o + 6)"-(a-6)''. 465. All the combinations of n letters, r together, repetitions permis- sible, are written down. Find the chance that a combination selected at random will contain no repetition. {n>r.) 466. All the combinations of n letters, r together, repetitions permis- sible, are written down. Find the chance that a combination selected at random will contain aE the n letters. («+x)[c+x)..,ton factors = a""'. 280 CHANCE. 478. If three numbers be named at random it is just as likely as not that every two of them will be greater than the third. 479. If the chance of any given person being ill at any given time is ff, what is the most likely population of a parish in which there are r persons HI? And what is the most likely number of persons to be ill at once in a parish of population n? 480. It is not known whether a set of dominoes goes to double six, double eight, or double nine. All three are equally likely, n dominoes are drawn from the set and contain no number higher than six. What are now the respective chances that the set goes to double six, double eight, or double nine? GEOUP XXV. (aEOMETEIOAL:) 481. A floor is paved with rectangular bricks each a inches long and b inches wide. A stick c inches long is thrown upon the floor so as to fall parallel to a diagonal of the bricks. Shew that the chance that it lies entirely on one brick is {a,^+V—c^-i-(a^+b'). 482. A circle, of diameter c, is thrown down on the same floor, shew that the chance that it lies on one brick is {a -c) (6 - c)-hab. 483. A ball of diameter c inches is thrown at random against a trellis composed of bars of width b inches crossing one another at right angles so as to leave openings of a inches square : shew that the chance of the ball passing clear through the treUis is {a-c)'-i-{a + b)^. 484. An indefinitely large piece of bread contains a spherical plimi of diameters. If the bread is cut up into cubical dice (each side=a>6), find the chance that the plum is cut. 485. A shot fired at a target of radius a makes a hole of radius b. If the centre of the shot is equally likely to hit all points of the target, find the chance that the hole made is completely within the circumference of the target. 486. If a point be taken at random on a finite straight line it is as likely as not that one of the parts into which it divides the line wiU be at least three times as great as the other. 487. If two points be taken at random on the circumference of a circle, it is as likely as not that one of the arcs into which they divide the circum- ference will be at least three times as great as the other. EXERCISES. 281 488. In either of the last two questions the ehanoe that one part should 2 be at least m times the other is = . m + 1 489. If two points be taken at random on the ciroumferenee of a circle, the odds are 2 to 1 that their (rectilinear) distance apart will be greater than the radius of the circle. 490. The odds are also 2 to 1 that their distance apart wUl be less than ^B times the radius of the circle. 491. Their distance is as likely as not to exceed ^^2 times the radius of the circle. 492. If a chord equal to the radius be drawn in a circle, and taken as the base of a triangle whose Tertes is a, point taken at random on the circum- ference, the area of the triangle is as likely as not to exceed the area of an equilateral triangle whose sides are equal to the radius. 493. The odds are 5 to 1 against the area of the triangle being more than double that of the equilateral triangle. 494. Two chords each equal to the radius are placed at random iri a circle so as not to intersect. If they become the opposite sides of a quadrilateral, the area of the quadrilateral is as likely as not to be more than 3 times as great as its least possible area. 495. A point is taken at random in each of two adjacent sides of a square. Shew that the average area of the triangle formed by joining them is one eighth of the area of the square. 496. If the points be taken on two adjacent sides of a triangle the average area wiU be one fourth of the area of the triangle. 497. If two points be taken at random on a finite straight line their average distance apart wUl be one third of the line. 498. On opposite sides of a rectangle two points are taken at random and are joined by a straight line, dividing the rectangle into two quadrilaterals. Find the chance that the area of one is more than n times the area of the other. 499. A coin whose diameter is one third of a side of a tile is thrown on the pavement of Question 154. Find the chance that it falls entirely on one tile. 500. On a chess-board, in which the' side of every square is a, there is thrown a coin of diameter 6, so as to be entirely on the board, which includes a border of width c outside the squares. Find the chance that the eoin is entirely on one square. (a>6>c.) 282 CHANCE. GEOUP XXVI. 501. A purse contains 2 sovereigns and 10 shillings. The coins are to be drawn out one by one and I am to receive aU the shillings that are drawn between the two sovereigns. What is my expectation ? 502. A die is thrown until every face has turned up at least once. Shew that on an average 14-^ throws will be required. 503. A pack of cards is repeatedly cut until at least one card of every suit has been exposed. Shew that on an average it must be out 10|- times. 504. Cards are drawn one by one from a pack until at least one card of every suit has been drawn. Shew that on an average 9444 cards must be drawn. 505. A deals from a pack of cards tiU he has dealt a spade : B takes the remainder of the pack and deals till he has dealt a spade : C takes the remainder and deals till he has dealt a spade. Shew that their expectations (of the number of cards they wiU deal) are equal. 506. A person draws cards one by one from a pack until he has drawn n hearts. Shew that his expectation is -^ re cards. 507. A person draws cards one by one from a pack untU he has drawn all the aces. How many cards may he expect to draw ? 508. A person draws cards one by one from a pack and replaces them till he has drawn two consecutive aces. How many cards may he expect to draw? 509. If he draw till he have drawn four consecutive spades how many cards may he expect to draw ? 510. One throws a die until one has thrown two consecutive sixes, and then receives a shilling for every six he has thrown. What is the value of his expectation ? 511. A pair of dice is to be thrown until doublets have turned up in every possible way. How many throws are to be expected ? 512. A bag contains m sovereigns and n shillings. A man is allowed to draw coins one by one until he has drawn p shillings. Shew that the value of his expectation is — £=- + ^ pounds. 513. What would be the expectation in the last question if each coin drawn were scored to the drawer's credit and replaced before the next drawing? 514. If he were to draw until he had drawn p pounds his expectation would be worth -^ + 20to shiUings. n+l EXERCISES. 283 515. If an experiment is equally likely to succeed or fail, the chance that it will succeed exactly n times in 2n trials is the ratio of the product of the first n odd numbers to the product of the first n even numbers. 516. If an experiment succeeds n+ 1 times for n times that it fails, the chance that it will succeed exactly «+ 1 times in 2»+ 1 trials is to the chance in the last question as {1- (2n+l)-'i}" to 1, or nearly as 4»+3 to 4n + 4. 517. If on an average 18 male children are bom for 17 female, find the chance that among the next 35 births there are exactly 18 males. 518. If the odds are m to n (where m>n) in favour of an experiment succeeding, the most likely number of successes in m+n trials is m; and m + 1 is a more likely number than m — 1 in the ratio of m« + m to mn + n ; but less likely than m in the ratio of m to ni+ 1. 519. If the chance of a trial succeeding is to its chance of failing as m : n, the most likely event in (m+n)r trials is mr successes and nr failures. 520. A candidate who correctly solves on an average » - 1 questions out of n goes in for an examination in two parts. In each part n questions are proposed and he is required to solve n - 1 of them. Shew that if the two parts were merged in one and he were required to solve 2 (m - 1) questions out of the 2n, his chance of passing would be improved in the ratio of 4 to 5 very nearly. [Accurately as 4m^ - 47i + 1 to Sn" - 5n + 1.] GEOUP XXVII. (Games.) 521. A r afflin g match is composed of five persons each throwing three times with seven coins, the one turning up the greatest number of heads to be the winner. A player having turned up 20 heads, it is required to find his chance of winning. 522. A and B play at a game which is such that no two consecutive games can be drawn. Their skill is such that when a game may be drawn the probability of it is q. Prove that the chance of r games being drawn o\itotpisCf-'^{l-qy-^{p-{q + l)r+l}^{^-2r+l). 523. In a game of mingled chance and skill, which cannot be drawn, the odds are 3 to 1 that any game will be decided by superiority of play and not by luck. A plays three games with B, and wins two. Prove that the odds are 3 to 1 in favour of A being the better player. If also B beat G two games out of three, prove that the chance of A winning the first three games he plays with C is ^^. 284 CHANCE. 524. My chance of muniug a game against A is a; against £ it is ^; against it is 7 ; and so on. One of these players in disguise enters into play with me and I win the first n games, shew that the chance of my win- ning the next game is (a"+i+^''+'+7"+i + ...)-i-(o"+j3^+7"+ ...). 525. At a chess tournament the players are supposed to be divisible into 2 classes, the odds on a member of the first in a game with a member of the second being 2 to 1. The second class is twice as numerous as the first. A player is observed to win a game, and a bet of 7 to 10 is made that he belongs to the first class. Shew that this is fair if there are 18 players. 526. A player has reckoned his chance of success in a game to be a, but he considers that there is an even chance that he has made an error in his caJcnlation affecting the result by p (either in excess or defect). Shew that this consideration does not affect his chance of success in a single game, but increases his chance of winning a series of games. 527. A and B play at draughts and bet £1 even on every game, the odds are 11, : 1 m favour of the player who has the first move. If it be agreed that the winner of each game have the first move in the next, shew that the advantage of having the first move in the first game is £/t-l. 528. Seven clubs compete annually for a challenge cup, which is to become the property of any club which wins it three consecutive years. Assuming aU the clubs to be equally good, what is the chance that last year's winner (not having won the previous year) will ultimately own the cup? 529. What is the chance of a club that has won the last two years? 530. If there be n clubs and the cup have to be won h years in succes- sion, what is the chance of a club that has won the last r years ultimately owning the cup? 531. Two players of equal skill engage in a match to play at draughts until one of them has won Ic games in succession. Shew that after A has won r games in succession, the odds are 2' +2' -2 to 2* -2' in favour of his winning the match. 532. A and B agree to play at chess until one has won k games more than the other. If they play equally well, when one has won r games more than the other, the odds are It + r : A; - r in favour of his winning the match. But if A plays /i times as well as B, then when A has won r games the odds are ij?' - pf'' : /i'"' - 1 in favour of his winning the match. EXERCISES. 285 533. A'e ohanoe of scoring any point being better than J5's chance in the ratio of five to four, A engages to score 14 in excess of B before B shall have scored three in excess of A. Shew that the odds are very slightly (about 265 : 263) in favour of A winning the match. 534. A and B play at a game which cannot be drawn, and the odds are seven to five in favour of A winning any assigned game. They agree to play until either A shall win by scoring 10 points ahead of B, or B shfffl win by scoring two points ahead of A. Shew that the odds are about 121 to 120 in favour of B winning the match. 535. A and B play at a game which cannot be drawn. A undertakes to win m games in excess of B, before B has won n in excess of A. Shew that, if the match is fair, ^'s chance of winning a single game must exceed JS's chance in the ratio : 1 very nearly. 536. A and B play a set of games, in which A'b chance of winning a single game is p, and B's chance q. Find (i) the chance that A wins m out of the first m+n, (ii) the chance that when A has won m games, m+n have been played, (iii) the chance that A wins m games before B wins n games. 537. A number of persons A, B, C, D ... play at a game, their chances of winning any particular game being o, j3, 7, 5 ... respectively. The match is won by j1 if he gains a games in succession ; by 5 if he gains 6 games in succession ; and so on. The play continues till one of these events happens. Shew that their chances of winning the match are proportional to l-a" ' l-;86 ■' *"• 538. A man possessed of a+ 1 pounds plays even wagers for a stake of 1 pound. Find the chance that he is ruined at the (a + 2x+ 1)"" wager and not before. 539. If a man playing for a constant stake, win 2re games and lose n games, the chance that he is never worse off than at the beginning and never better off than at the end is (n^ + n+2)-i-(im'+ 6«+2). 540. If he win 2n + 1 games and lose n+\ games, the chance is n-=-{4«+6). 286 CHANCE. GROUP XXVm. (Expectation or Life.) Questions of Present YaVue are to be solved approximately , interest being calculated at 5 per cent. 541. Prove that at birth the expectation of life is 542. Prove that if E be the expectation at birth, the expectation at the age of a; years willbe^ + (i;-JBj^.i-B;t4^-Bj^^3-...)-7-iJ-5. 543. Thirteen persons meet at dinner, their ages being n, m + 1, m+2, and so on, to »+ 12. Shew that the chance that they will all be alive a year hence is JJ,^i3-r-E„, and express the chance that they will all be alive five years hence. 544. Three persons bom on the same day keep their 21st birthday together. They agree to keep in like maimer their 31st, 4lEt, Slst, &c., as long as they are all alive. How many such celebrations are there Kkely to be? 545. An aunt aged 70 has a nephew bom to-day and she promises to make him three annual payments of £100 commencing on his 15th birthday, subject to their both being alive. What is the chance that the nephew will be entitled to aU the 3 payments? And what is the chance that he will receive none of them? 546. What is the present value of the aunt's gift? 547. If aunt and nephew are both aUve 5 years hence what wUl then be the value of the nephew's expectation? 548. If out of 100 persons who reach the age of 87 years, 19 die in the first year, 17 the next, year, 15 the next and so on in a. p., shew that the expectation of life at 87 is 3-35 years, at 88 it is 3-0i85, and at 89 it is 2-6875. 549. On the hypothesis of the last question what is the present value of an annuity of £100 per annum on the life of a person aged 87: the first payment being due a year hence? 550. On the same hypothesis what annual premium should a person pay for an insurance of £1000, the first payment being made on his 87th birthday? 551. A who is aged 87 is entitled to the reversion of an annuity of £100 a year on his own life after the death of S who is just three years older than himself. What is the present value of 4's expectation on the hypothesis of the last three questions? EXEKCISES. 287 552. What is the value of his expectation on the hypothesis that out of 9 persons who reach the age of 87, two die in the first year and one every year afterwards till all are dead? 553. If (on another planet) the conditions of life are such that out of every 95 persons who reach the age of 5 years one dies every year till all are dead, what is the chance that a person aged m years will outlive another person whose age is n years ? («>m>5.) 554. If out of n persons who reach the age of N years one dies every year till all are dead, shew that the expectation of life at the age of N+r is i(B-r). 555. On the hypothesis of the last question the chance that three persons aged N, N+a, N+2a respectively should all be alive a years hence is {n-3a)-i-n. 556. On the same hypothesis find the present value of an annuity on the Ufe of a person aged N years, £t being the interest on £1 for 1 year. 557. On the same hypothesis find the present value of an annuity to continue during the joint lives of two persons now aged N and N+ a. 558. What would be the present value of the annuity in the last question if it were to continue until both persons were dead ? 559. If out of m* persons who reach the age of N years, 2» - 1 die in the first year, 2tt-3 in the second, 2n — 5 in the third, and so on, shew that the expectation of life at the age of ^+r is —^ — i--^. . "^ 3 6(»— r) 560. On the hypothesis of the last question, shew that the chance that p persons whose ages are N, N+ 1, N+ 2, ... N+p - 1, will live another year is 1 1 - - J , and the chance that they will all live for r years is { C> "^-v- C? }^. GEOUP XXIX. (Balls from Bags, &o.) 561. A bag contains 2m balls of which 2n are white. A second contains 3m balls of which 3« are white. If two balls are to be drawn from each bag, which bag is the more likely to give both white? and which is the more likely to give at least one white? 562. Each of two bags contains m sovereigns and n shillings. If a man draw a coin out of each bag he is more likely to draw two sovereigns than if all the coins were in one bag and he drew two. 563. There are 3 balls in a bag, and each of them may with equal proba- bility be white, black, and red. A person puts in his hand and draws a ball. It is white. It is then replaced. Find the chance (i) of aU the baUs being white, (ii) that the next drawing will give a red ball. 288 CHANCE. 564. A bag contains a black balls and h white balls. A second bag contains a+c black balls and 6-c white balls, m balls are drawn from one and n from the other, and are found to be all black. Shew that the odds are <^»,o-».: y, both being even or both odd, so that x+y=2s ; and if x be expressed as the sum of y integers, aU the ways of so expressing it being equally likely, the chance of aU the y integers being odd is nj-hllj. (Notation of page 101.) EXERCISES. 293 618. A pack of cards consists of ^ suits of g cards each, numbered from 1 up to g. A card is drawn and turned up : and r other cards are drawn at random. Find the chance that the card first drawn is the highest of its suit among all the cards drawn. 619. A pack of n different cards is laid face downwards on «, table. A person names a certain card. That and aU the cards above it are shewn to him, and removed. He names another; and the process is repeated until there are no cards left. Find the chance that, in the course of the operation, 'a card was named which was (at the time) at the top of the 620. If the birth rate is double the death rate, shew that the annual multiplier is given by the equation (Notation of p. 196, S„ denoting 1 - iJ„. ) GEoup xxxn. 621. A man writes a number of letters greater than eight, and directs the same number of envelopes. If he puts one letter into each envelope at random, the chance that all go wrong is (to six decimal places) -367879. 622. A class of twelve men is arranged in order of merit: find the chance that no name is in the place it would have occupied if the class had been in alphabetical order. 623. A list is to be published in three classes. The odds are m to 1 that the examiners wiU decide to arrange each class in order of merit, but if they are not so arranged, the names in each will be arranged in alphabetical order. The list appears, and the names in each class are observed to be in alphabetical order, the numbers in the several classes being a, b, and c. What is the chance that the order in each class is also the order of merit? 624. A goes to haU p times in q consecutive days and sees B there r times. What is the most probable number of times that B was in hall in the q days ? Ex. Supposep=4, g=7, r=S. 625. If n witnesses concur in reporting an event of which they received information from another person, the chance that the report is true will be (p»^'- + 5r^i)^(jp"+g'') wherep=l-3 is the chance of the correctness of a report made by any single person. 294 CHANCE. 626. A judges correctly 5 times out of 6, and B 3 times out of 4. The same four £5 notes have been submitted to them both independently. A declares that 3 of the notes are good without specifying them. B declares that only one is good. How much ought one to give for the notes ? 627. There are two clerks in an office, each of whom goes out for an hour in the afternoon, one may start at any time between two and three o'clock, the other at any time between three and four, and all times within these limits are equally likely. Find the chance that they are not out together. 628. The reserved seats in a concert-room are numbered consecutively from 1 to m+m+r. I send for m consecutive tickets for one concert and n consecutive tickets for another concert. What is the chance that I shall find no number common to the two sets of tickets? 629. Two persons are known to have passed over the same route in opposite directions within a period of time m+n+r, the one occupying time m, and the other time n: find the chance that they will have met. 630. A writes to B requiring an answer within n days. It is known that B will be at the address on some one of these days, any one equally likely. It is a y-days' post between A and B. If one in every q letters is lost in transit, find the chance that A receives an answer in time. (M>2p.) 631. There are n vessels containing wine, and n vessels containing water. Each vessel is known to hold a, a + 1, a +2, ... or a+m-1 gallons. Find the chance that the mixture formed from them aU will contain just as much wine as water. 632. A vessel is fiUed with three liquids whose specific gravities in descending order of magnitude are Sj, 8^ Sy All volumes of the several liquids being equally likely, prove that the chance of the specific gravity of the mixture being greater than S is (Si -SJ (Si -S3)' ""^ (Sa- S3) (Si -S3)' according as S lies between Sj and 8^, or between Sj and S3. 633. A man drinks in random order n glasses of wine and n glasses of water (all equal) ; shew that the odds are ?i to 1 against his never having drunk throughout the process more wine than water. 634. If n men and their wives go over a bridge in single file, in random order, subject only to the condition that there are to be never more men than women gone over, prove that the chance that no man goes over before his wife is {n+ 1) 2~". EXEECISES. 295 636. A dinner party consists of n gentlemen and their wives. When « - 1 gentlemen have taken m - 1 ladies down to dinner, no gentleman taking his own wife, and aJl possible arrangements being equally likely, the odds are ||« to |[m-l against the gentleman and lady remaining being husband and wife. 636. A coin is tossed m+n times, where m>re. Shew that the chance of there being at least m consecutive heads is (n+2)-=-2"'+i. 637. If a coin be tossed n times, the chance that there will not be k con- secutive heads is the coe£Gicient of x" in the expansion of 1^ 1 +x+t'+ ... +x'~^ _ 2"' l-a;-x2-...-a;« 638. A person throws up a coin n times : for every sequence of m throws, heads or taUs, for all possible values of m, he is to receive 2™ - 1 shillings ; prove that the value of his expeotaion is 3» (n+ 3) pence. 639. A having a single penny, throws for a stake of a penny with B, who has at least 8 pence. Shew that the chance that A loses his peimy at the 17th throw and not before is 1430 h-21'. 640. If a player repeatedly stake the same sum in a, series of either 2n or 2» - 1 even wagers, the chance that throughout the play he is never worse off than he was at the beginning is expressed by the ratio of the product of the first n odd numbers to the product of the first n even numbers. ANSWERS TO THE EXERCISES. Gboup I. 1. 15. 2. 10000. 3. 20. 4. 8. 5. 9. e. 48. 7. 25. 8. 480, 437. 9. 60. lO. 1080. 11. 30. 12. 60. 13. 147. 14. 134. 15. 143. 16. 110 > 108. 17. 40320,_5040. 18. 480, 22. 19. 10. 20. 60. Gbouf II. 21. 20. 22. 70. 23. 432. 24. 26820600. 25. 3. 26. 360, 120, 24. 27. 30. 28. 729. 29. 63. 30. 756. 31. 60. 32. 24. 33. 18|60-f-|20[40, 3|60-=-[20|40. 34. 675675. 35. 118755. 37. 10156250. 38. 1728. 39. 55440, 174240. 40. |32-^|12|12|8. 296 ANSWERS. Gboup III. 41. 166320. 42. 3360. 44. 780. 45. 6. 46. 6. 47. 720. 48. 90720. 49. 90. SO. 3386880. 51. 2880. 62. 3023. S3. 332639. 54. 277199. 55. 144. 56. 1058399. S7. 63, 18. 58. 96. 59. 12. 60. 112266000. Geoup IV. 61. 945. 62. 945. 63. 280. 64. 10716300000. 65. 21432600. 66. 264600. 67. 924. 68. 9129120. 69. |20. 70. 9071. 71. 1919. 72. 43. 73. 183. 75. 6. 76. 12. 78. 5849. 79. 690. 80. 21461. Oboup Y. 81. 35. 82. 8820. 83. 900. 84. 20. 85. 2. 86. 126. 87. 56. 88. 84. 89. 30. 90. 6, 54. 91. 102. 92. 265. 93. 548. 94. 17689. 95. 4020. 96. 428. 97. 864. 98. 2220. 99. 88080. lOO. 15240. Geoup VI. lOl. 100. 102. 2030. 103. 231. 104. 1286. 105. 90000, 3125, 2500, 1024, 768, 120, 96. 106. 21. 107. 56. 108. 293930, 24310, 45. 109. (|13)S (|13)5-=-2«.3i8. IIO. (|13)«v 213.3''. 111. 5(17-4-212.32, 6 117-=- 215.35. 112. 4782968. 114. 2815. 115. 7405. 116. 14308. 117. 1228613679150. 119. 40824. 120. 53922. GnonpVII. 121. 5. 122. 4,5. 123. y=mx=('nm-m!')-i-(m^-n). 134. j7i(3?i2-3ra + 2). 135. \pq + r -i-{q + l)'(\qy. Gboup VXH. 141. 20. 142. 20. 143. 220. 144. 6. 145. 4. 146. Gi-G^. 147. ipq(P + l.i- 368. A. i4. /^- 369. As 11 : 10 : 9 : 9. 370. As 16 : 12 : 9. 371. 95 to 6. 372. 9 to 8. 373. 12 to 7. 374. J. 375. J. 376. f. 377. J^. 378. i,i, J. 379. A, A. A- Group XX. 383. (i) 13' ^ 17 . 25 . 49. (ii) 48 . 44 . 40 H- 51 . 50 . 49. (iii) 36.22. 10 H- 51. 50. 49. 384. 6 . 12*. 13'--?^. 385. J. 386. JliJ. 387. As 773347 : 399303 : 378622. 388. |f , if, f |f . 391. 154 to 153. 392. im< ^i- 393. i, m. 394. JlJf, |4|f- 395. The latter. 396. 22 shillings. 397. 21 ± 19(-8)'» crowns. 398. ^. 400. iV i Vs Va Vo - w. 20 298 ANSWERS. Group XXI. 401. ff- ^O^. ^^. 403. -g,^. 404. Iff. 40S. 2-96. 406. 3-19. 408. Approximately £213. 409. Approximately £127. 10s. 410. 3-35. 411. As 24649 : 10286. 415. 245 to 124. 416. if«|. 417. ^Hi^. 418. im^- 419. 1^. 420. m^ GBonp XXII. 421. ^fe. 422. r^. 423. |f|i. 424. ill. 425. jffj. 426. -jj^ir- 437. j^ff^-- 428. /A'A- 429. -^m^. 430. yifflTr- 431- 432. rtWAj- *38- liVaVas - 434- tMt- 435. i^^. 436. ,^^. 437. -^iU- 438. -j^It- GBonp XXIII. 443. A. U. A- 444. i|, i^, :^. 456. H|. Group XXIV. 465. Cj-^BJ. 466. C;-HiJ;. 467. |y-m |roTO-H|iy(|TC)'" where N=G^. 468. -367879. 469. 1 59 Zf 59^-^ 60^ |x 1 60 - x , 60|60|88-H |29|119. 471. 8. 472. 15i, 54|. 475? 12{n-l)-T-(3™-l)(BK-2). 479. Greatest integer in r-i-d and in fl (m + 1). 480. Inversely as C^ : C« : CJf. Group XXV. 484. X-^a-hf-i-aK 485. (a-iy^a^. 498. 4n-=-(m+l)2. 499. {1Z-&JZ)^%. 500. ( + c)-=-{(o+l)(6 + l)(c + l)-li. 592. 3(37i«-37i+2)-r(TCS + 37S« + 2B). 593. 72 h- (n + 2)(n2 + 4n + 6). 594. 9(n + 3)^5(n2 + 4m + 6). 595. i{n + 3). 596. i6(n + 2). GkoupXXXI. 601. ^1-iy. 602. fl--y. 605, 606. (i]*. 608. QV. 609. f^y. 610. Qy. 613. 4319 days 18 min. Group XXXII. 622. -367879. 623. (m + l)-r (m+[a[6|c. 624. (3 + l)r-^ 2) or next lower integer. 626. £10. 16s. 10|d. nearly. 627. i. 628. (r + l)(r+2)-l-(m+r + l)(n + r + l). 629. (iim+mr+»r)-=-(m+r)(7S+r). 630. (n - 2p)(g - 1)' -f nj'. 631. Middle coefficient in {(l-a:")^-m(l-a!)r. CAMBBIDOE: PBINTED by C. j. CLi.Y, M.A. AND SONS, AT THE UNIVEBSIIY PBESS. S^tember 1893. 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