# Page:EB1911 - Volume 22.djvu/396

382
[METHODS OF CALCULATION
PROBABILITY

at an examination and r optional subjects from which each candidate chooses one (r > s), what is the probability that no two candidates should choose the same subject? If the candidates be arranged in any order, the probability that the second candidate should not choose the same subject as the first candidate is (n - 1)/n. The probability that the third candidate will not choose either of the two subjects taken by the aforesaid candidates is (n - 2)/n, and so on. Thus the required probability is

n(n − 1)(n − 2) ⋅ ⋅ ⋅ {n − (s − 1)}/ns.

38. When as in these cases the interest of the problem lies chiefly in the application of the theory of combinations, or permutations, there is a propriety in Whitworth's enunciation of the questions under the head of choice rather than chance. It comes to the same whether we say that there are x ways in which an event may happen, or that the probability of its happening in an assigned one of those ways is 1/x. For example, suppose that there are n couples waltzing at a ball; if the names of the men are arranged in alphabetical order, what is the probability that the names of their partners will also be in alphabetical order? The probability that the man who is first in alphabetical order should have for partner the lady who is first in that order is 1/n. The probability that the man who is second alphabetical order should have for partner the lady who is second in that order is 1/(n - 1), and so on. Therefore the required probability is 1/n!. Or it may be easier to say that the number of ways, each consisting of a set of couples in which the party can be arranged, is n!; of which only one is favourable.

39. The same principle governs the following question. For how many days can a family of 10 continue to sit down to dinner in a different order each day; it not being indifferent who sits at the head of the table—what is the absolute, as well as the relative, position of the members? The number of permutations, viz. 10!, is the answer. If we are to attend to the relative position only—as would be natural if the question related to 10 children turning round a flypole—the number of different arrangements would be only 9!

40. Method of Equations in Finite Differences.—The last question may serve to introduce a method which Laplace has applied with great éclat to problems in probabilities. Let yn be the number of ways in which n men can take their places at a round table, without respect to their absolute position; and consider how the number will be increased by introducing an additional man. From every particular arrangement of the original n men can now be obtained n different arrangements of the n + 1 men (since the additional man may sit between any two of the party of n). Hence yn+1 = nyn, an equation of differences of which the solution is C (n - 1)! The constant may be determined by considering the case in which n is 2.

41. The following example is not quite so simple. If a coin is thrown n times, what is the chance that head occurs at least twice running? Calling each sequence of n throws a “case,” consider the number of cases in which head never occurs twice running; let un be this number, then 2nun must be the number of cases when head occurs at least twice successively. Consider the value of un+1; if the last or (n + 2)th throw be tail, un+2 includes all the cases (un+1) of the n + 1 preceding throws which gave no succession of heads; and if the last be head the last but one must be tail, and these two may be preceded by any one of the un favourable cases for the first n throws. Consequently,

un+2 = un+1 + un

If α, β are the roots of the quadratic x2x − 1 = 0, this equation gives[1]

un = Aαn + Bβn.

Here A and B are easily found from the conditions u1 = 2, u2 = 3; viz.

${\displaystyle {\text{A}}={\frac {\alpha ^{2}}{\alpha -\beta }}}$, ${\displaystyle {\text{B}}={\frac {\beta ^{2}}{\beta -\alpha }}}$,

whence ${\displaystyle u_{n}={\frac {n+2}{2^{n+1}}}\{1+{\frac {(n+1)n}{1\cdot 2\cdot 3}}5+{\mbox{etc.}}\}}$.

The probability that head never turns up twice running is found by dividing this by 2n, the whole number of cases. This probability, of course, becomes smaller and smaller as the number of trials (n) is increased. This is a particular case of a more general problem solved by Laplace[2] as to the occurrence i times running of an event of which the probability at one trial is p.

42. In such problems where we now employ the calculus of finite difference Laplace employed his method of generating functions. A distinguished instance is afforded by the problem of points which was put by the Chevalier de Méré to Pascal and has exercised generations of mathematicians. It is thus stated by Laplace.[3] Two players of equal skill have staked equal sums; the stakes to belong to the player who shall have won a certain number of games. Suppose they agree to leave off playing when one player, A, wants x “points” (games to be won) in order to complete the assigned number, while the second player wants x′ points: how ought they to divide the stakes? This is a question in Expectation, but its difficulty consists in determining the probability that one of the players, say A, shall win the stakes. Let that probability be yx,x′. Then, after the next game, if A has won, the probability of his winning the stakes will be yx-1,x′. But if A loses, B winning, the probability will be yx,x′-1. But these alternatives are equally likely. Accordingly the probability of A winning the stakes may be written

½yx-1,x′ + ½yx,x′-1.

This is the same probability as that which was before written yx,x′. Equating the two expressions we have, for the function y, an equation of finite difference involving two variables, of which the solution is[4]

${\displaystyle y={\frac {1}{2}}x\{1+{\frac {x}{1}}{\frac {1}{2}}+{\frac {x(x+1)}{1\cdot 2}}{\frac {1}{2^{2}}}\}+\cdots +{\frac {x(x+1)\cdots (x+x'-2)}{1\cdot 2\cdots (x'-1)}}{\frac {1}{2^{x'-1}}}\}}$.

43. The problem of points is to be distinguished from another classical problem, relating to a contest in which the winner has not simply to win a certain number of games, but to win a certain number of counters from his opponent.[5] Space does not admit even the enunciation of other complicated problems to which Laplace has applied the method of generating functions.

44. Probability of Causes Deduced from Observed Events.—Problems relating to the probability of alternative causes, deduced from observed effects, are usually placed in the separate category of “inverse” probability, though, as above remarked,[6] they do not necessarily involve different principles. The difference principally consists in the need of evidence, other than that which is afforded by the observed event, as to the probability of the alternative causes existing and operating. The following is an example free from the difficulty incident to unverified a priori probabilities, which commonly besets this kind of problem. A digit having been taken at random from mathematical tables (or the expansion of an endless constant such as π); a second digit is obtained by taking from a random succession of digits one that added to the first digit makes a sum greater than 9. Given a result thus formed, what are the respective probabilities that the second digit should have been 0, 1, 2, . . . 8 or 9? In the long run the first digit assumes with equal frequency the values 0, 1, 2 . . . 8, 9. Accordingly the second digit can never be 0. There is only one chance of its being 1, namely when the first digit is 9. If the second digit is 2, and the first either 8 or 9, the observed effect will be produced. And so on. If the second digit is 9, the effect may occur in nine ways. Accordingly in the long run of pairs thus formed it will occur that the cases or causes which are defined by the circumstances that the second digit is 0, 1, 2, . . . 8, 9, respectively, will occur with frequencies in the following ratios 0 : 1 : 2 . . . 8 : 9. The probability of the observed event having been caused by a particular (second) digit, e.g. 7, is 7/(0 + 1 + 2 + . . + 9) = 7/45.

45. The following example taken from Laplace[7] is of a more familiar type. An urn is known to contain three balls made up of white and black balls in some unknown proportion. From this urn a ball is extracted m times (being each time replaced after extraction). If a white ball is drawn every time, what are the respective probabilities that the number of white balls in the urn are 3, 2, 1 or 0? By parity of reasoning it appears that in the first case the result is certain, its probability 1, in the second case the probability of the observed event occurring is (⅔)m, in the third case that probability is (⅓)m, in the fourth case zero. Accordingly the respective inverse probabilities are in the ratios

1 : (⅔)m : (⅓)m : 0;

provided that (as in the preceding example, with respect to the second digits) the alternative causes, the four possible constitutions of the urn, are (a priori) equally probable. This is rather a bold assumption with respect to the contents of concrete urns[8] and similar groupings; but with regard to things in general may perhaps be justified on the principle of cross-series.[9]

46. Often in the investigation of causes we are not thrown back on unverified a priori probabilities. We have some specific evidence though of a very rough character. An example has been cited from Mill in a preceding paragraph.[10] Against the improbabilities calculated by the methods of the present section there has often to be balanced an improbability evidenced by common sense, which does not admit of mathematical calculation. Bertrand[11] puts the following case. The manager of a gambling house has purchased a roulette table which is found to give red 5300 times, black 4700 times, out of 10,000 trials. The purchaser claims an indemnity from the maker. What can the calculus tell us as to the justice of the claim? Nothing

1. Cf. Boole's Finite Differences, ch. vii. § 5.
2. Op. cit. liv. II. ch. ii., No. 12.
3. Op. cit. liv. II. ch. ii., No. 8.
4. A clear and corrected version of Laplace's reasoning is given by Todhunter, History. . . of Probability, art. 973, p. 528, with reference to the more general cases in which the “skills” of each party—their chances of winning a single game—are not equal but respectively p and q (p + q = 1). See also Czuber, Wahrscheinlichkeitstheorie, pp. 30 seq.
5. See Todhunter, op. cit. art. 107, and other articles referring to duration of play. See also Boole, Finite Diferences, ch. xiv., art. 7, ex. 6.
6. Above, par. 13.
7. Op. cit. liv. II. ch. i. No. 1.
8. Cf. Bertrand, op. cit. § 118.
9. Above, par. 5.
10. Par. 13.
11. Op. cit. § 134.