Page:EB1911 - Volume 22.djvu/398

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Thus each judge implicitly assigns the probabilities

1/n2, 1/n(1/n + 1/n − 1), 1/n(1/n + 1/n − 1 + 1/n − 2),

to the causes as they stand on his list, beginning from the lowest. The values asiigned for the probability of each alternative cause may be treat as so many equally authoritative observations representing a quantity which it is required to determine. According to a general rule given below[1] the observations are to be added and divided by their number; but here if we are concerned only with the relative magnitudes of the probabilities in favour of each alternative it suffices to compare the sums of the observations. We thus arrive at Laplace's rule. Add the numbers found on the different lists for the cause A, for the cause B, and so on; that cause which has the greatest sum is the most probable.

53. Probability of Future Effects deduced from Causes.—Another class of problems which it is usual to place in a separate category are those which require that, having ascended from an observed event to probable causes, we should descend to the probability of collateral effects. But no new principle is involved in such problems. The reason may be illustrated by the following modification of the problem about digits which was above set[2] to illustrate the method of deducing the probability of alternative causes. What is the probability that if to the second digit which contributed to the effect there described there is added a third digit taken at random, the sum of the second and third will be greater than 10 (or any other assigned figure)? The probabilities—the a posteriori probabilities derived from the observed event (that the sum of the first and second digit exceeds 9)—each multiplied by 45, of the alternatives constituted by the different values 0, 1, 2, . . . 8, 9 of the second figure are written in the first of the subjoined rows.

0 1 2 3  4  5  6  7  8  9
0 0 1 2  3  4  5  6  7  8
0 0 2 6 12 20 30 42 56 72

Below each of these probabilities is written the probability, × 10 that if the corresponding cause existed the effect under consideration would result. The product of the two probabilities pertaining to each alternative way of producing the event gives the probability of the event occurring in that way. The sum of these products which are written in the third row divided by 45 × 10, viz. 240/450 = 8/15, is the required probability. It may be expected that actual trial would verify this result.

54. “Rule of Succession.”—One case of inferred future effects, sometimes called the “rule of succession,” claims special notice as having been thought to furnish a test for the cogency of induction. A white ball has been extracted (with replacement after extraction) n times from an immense number of black and white balls mixed in some unknown proportion; what is the probability that at the (n + 1)th trial a white ball will be drawn.? It is assumed that each constitution of the mélange[3] formed by the proportion of white balls (the probability of drawing a white ball), say p, is a priori as likely to have any one value as another of the series

Δp, 2Δp, 3Δp, . . . 1 - 2Δp, 1 - Δp, 1.

Whence a posteriori the probability of any particular value of p as the cause of the observed recurrence is pn/∑pn, where p in the denominator receives every value from Δp to 1. The probability that this cause, if it exists, will produce the effect in question, the extraction of a white ball at the (n + 1)th trial, is p. The probability of the event, obtained by summing the probabilities of all the different ways in which it may occur, is accordingly ∑pn+1/∑pn, where p both in the numerator and the denominator is to receive all possible values between Δp and 1. In the limit we have


In particular if n = 1, the probability that an event which has been observed once will recur on a second trial is ⅔. These results are perhaps not so absurd as they have seemed to some critics, when the principle of “cross-series”[4] is taken into account. Among authorities who seem to attach importance to the rule of succession, in addition to the classical writers on Probabilities, may be mentioned Lotze[5] and Karl Pearson.[6]

Section III.—Calculation of Expectation.

55. Analogues of Preceding Problems.—This section presents problems analogous to the preceding. If n balls are extracted from an urn containing black and white balls mixed up in the proportions p : (1 - p), each ball being replaced after extraction, the expected number of white balls in the set of n is by definition np.[7] It may be instructive to verify the consistency of first principles by demonstrating this axiomatic proposition.[8] Consider the respective probabilities that in the series of n trials there will occur no white balls, exactly one white ball, exactly two white balls, and so on, as shown in the following scheme:—

No. of white balls 0, 1, 2,  . . . n
Corresponding probability   (1 − p)n,   n!/(n − 1)!1(1 − p)n−1p,   n!/(n − 2)!2!(1 − p)n−2p2   . . . pn

To calculate the expectation of white balls it is proper to multiply 1 by the probability that exactly one white ball will occur, 2 by the probability of two white balls, and so on. We have thus for the required expectation

n!/(n − 1)!(1 − p)n−1p + n!/(n − 2)!(1 − p)n−2p2 + . . . + n!/(nr)!(r − 1)!(1 − p)nrpr + . . . + npn

= np[(1 − p)n−1 + (n − 1)(1 − p)n−2p + . . . + (n − 1)!/(nr)!(r − 1)!(1 − p)nrpr−1 + . . . + pn−1]

= np[(1 − p) + p]n−1 = np.

The expectation in the case where the balls are not replaced—not similarly axiomatic—may be found by approximative formulae.[9]

56. Games of Chance.—With reference to the topic which occurred next under the head of probabilities, a distinction must be drawn between the number of trials which make it an even chance that all the faces of a die will not have turned up at least once, and the number of trials which are made on an average before that event occurs. We may pass from the probability to expectation in such cases by means of the following theorem. If s is the number of trials in which on an average success (such as turning up every face of a die at least once) is obtained, then s = 1 + f1 + f2 + . . . ; where fr denotes the probability of failing in the first r trials. For the required expectation is equal to 1 × probability of succeeding at the first trial + 2 × probability of succeeding at the second trial + &c. Now the probability of succeeding at the first trial is 1 − f1; the probability of succeeding at the second trial, (after failing at the first) is f1(1 − f2); the probability of succeeding at the third trial is similarly f2(1 − f3); and so on. Substituting these values for the expression for the expectation, we have the proposition which was to be proved. In the proposed problem

fn = 6(5/6)n − 15(4/6)n × 20(3/6)n − 15(2/6)n + 6(1n/6)

Assigning to n in each of these terms, every value from 1 to ∞ we have 6⋅5/6/(1 − 5/6), = 30, for the sum of the first set, with corresponding expressions for the sets formed from the following terms. Whence {{{1}}} By parity of reasoning it is proved that on an average 7419/630 cards[10] must be dealt before at least one card of every suit has turned up.[11]

57. Dominoes are taken at random (with replacement after each extraction) from the set of the kind described in a preceding paragraph.[12] What is the difference (irrespective of sign) to be expected between the two numbers on each domino? The digit 9, according as it is combined with itself, or any smaller digit, gives the sum of differences

0 + 1 + 2 + . . . + 9.

The digit 8 combined with itself or any smaller digit gives the sum of differences 0 + 1 + 2 + . . . + 8 and so on. The sum of the differences is ∑1/2r. r + 1, where r has every integer value from 1 to 9 inclusive, = 9(9 + 1)(9 + 2)/2⋅3, = 165. And the number of, the differences is 10 + 9 + 8 + . . . + 2 + 1 = 55. Therefore the required expectation is 165/55 = 3.

58. Digits taken at Random.—The last question is to be distinguished from the following. What is the difference (irrespective of sign) between two digits, taken at random from mathematical tables, or the expansion of an endless constant like π? The combinations of different digits will now occur twice as often as the repetitions of the same digit. The sum of the differences may now be obtained from the consideration that the sum of the positive differences must be equal to sum of the negative differences when the null differences are distributed equally between the positive and the negative set. The sum of the positive set is, as before,

  1. Below, pars. 135, 136. A difficulty raised by Cournot with respect to the determination of several quantities which are connected by an equation does not here arise. The system of values determined for the several causes fulfils by construction the condition that the sum of the values should be equal to unity.
  2. Above, par. 44.
  3. It comes to the same to suppose the total number of balls in the mixture to be N; and to assume that the number of white balls is a priori equally likely to have any one of the values 1, 2, . . . N - 1, N.
  4. Above, par. 5.
  5. Logic, bk. ii. ch. ix. § 5.
  6. Grammar of Science, ch. iv. § 16. Cf. the article in Mind above referred to, ix. 234.
  7. See the introductory remarks headed “Description and Division of the Subject.”
  8. Cf. above, par. 25.
  9. See Pearson, Phil. Trans. (1895), A.
  10. Whitworth, Exercises, No. 502.
  11. Ibid. No. 504, cf. above, par. 29.
  12. Ibid. par. 36.