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52, pn for 13, qn for 39 (where p + q = 1; pn and qn are integers). A formula thus generalized is proposed by Professor Karl Pearson^[1] as proper to represent the frequency with which different values are assumed by a quantity depending on causes which are not independent.

29. Miscellaneous Examples: Games of Chance.—The majority of the problems under this heading cannot, like the preceding two, be regarded as conducing directly to statistical methods which are required in investigating some parts of nature. They are at best elegant exercises in a kind of mathematical reasoning which is required in most of such methods. Games of chance present some of the best examples. We may begin with one of the oldest, the problem which the Chevalier de Méré put to Pascal when he questioned: How many times must a pair of dice be thrown in order that it may be an even chance that double six—the event called sonnez—may oocur at least once?^[2] The answer may be obtained by finding a general expression for the rob ability that the event will occur at least once in n trials; and then determining n so that this expression = ½. The probability of the event occurring is the difference between unity and the probability of its failing. Now the probability of “sonnez” failing at a single throw (of two dice) is 35/36. Therefore the probability of its failing in n throws is $({\frac {35}{36}})^{n}$. Whence we obtain, to determine n, the equation $1-({\frac {35}{36}})^{n}={\frac {1}{2}}$, which gives n = 24.605 nearly.

30. In the preceding problem the quaesitum was (unity minus) the probability that out of all the possible events an assigned one (“sonnez”) should fail to occur in the course of n trials. In the following problem the quaesitum is the probability that out of all the possible events one or other should fail—that they should not all be represented in the course of n trials. A die being thrown n times, what is the probability that all three of the following events will not be represented (that one or other of the three will not occur at least once); viz. (a) either ace or deuce turning up, (b) either 3 or 4, (c) either 5 or 6. The number of cases in which one at least of these events fails to occur is equal to the number of cases in which (a) fails, plus the number in which (b) fails, plus the number in which (c) fails, minus the number of cases in which two of the events fail concurrently (which cases without this subtraction would be counted twice).^[3] Now the number of cases in which (a) fails to occur in the course of the n trials is $({\frac {2}{3}})^{n}$ of all the possible cases numbering 3ⁿ. Like propositions are true of (b) and (c). The number of cases in which both (a) and (b) fail is $({\frac {1}{3}})^{n}$ of the total;^[4] and the like is true of the cases in which both (a) and (c) fail and the cases in which both (b) and (c) fail. Accordingly the probability that one at least of the events will fail to occur in the course of n trials is

$3({\frac {2}{3}})^{n}-3({\frac {1}{3}})^{n}$.

31. One more step is required by the following problem: If n cards are dealt from a pack, each card after it has been dealt being returned to the pack, which is then reshuffled, what is the probability that one or other of the four suits will not be represented? The probability that hearts will fail to occur in the course of the n deals is $({\frac {3}{4}})^{n}$; and the like is true of the three other suits. From the sum of these probabilities is to be subtracted the sum of the probabilities that there will be concurrent failures of any two suits; but from this subtrahend are to be subtracted the proportional number of cases in which there are concurrent failures of any three suits (otherwise cases such as that in which e.g. hearts, diamonds and clubs concurrently failed^[5] would not be represented at all). Now the probability of any assigned two suits failing is $({\frac {2}{4}})^{n}$; the probability of any assigned three suits failing is $({\frac {1}{4}})^{n}$. Accordingly the required probability is

$4({\frac {3}{4}})^{n}-6({\frac {2}{4}})^{n}+4({\frac {1}{4}})^{n}$.

The analogy of the Binomial Theorem supplies the clue to the solution of the general problem of which the following is an example. If a die is thrown n times the probability that every face will have turned up at least once is^[6]

$1-6({\frac {5}{6}})^{n}+15({\frac {4}{6}})^{n}-20({\frac {3}{6}})^{n}+15({\frac {2}{6}})^{n}-6({\frac {1}{6}})^{n}$.

32. If in the (first) problem stated in paragraph 31 the cards are dealt in the ordinary way (without replacement), we must substitute for $({\frac {3}{4}})^{n}$, the continued product${\frac {39}{52}}\cdot {\frac {38}{51}}\cdots {\frac {39-(n-1)}{52-(n-1)}}$; for $({\frac {2}{4}})^{n}$ the continued product ${\frac {26}{52}}\cdot {\frac {25}{51}}\cdots {\frac {26-(n-1)}{52-(n-1)}}$, and so on.

33. Still considering miscellaneous examples relating to games of chance let us inquire what is the probability that at whist each of the two parties should have two honours?^[7] If the turned-up card is an honour, the probability that of the three other honours an assigned one is among the twenty-five which are in the hands of the dealer or his partner, while the remaining two honours are in the hands of the other party, is 25/51⋅26/50⋅25/49. But the assigned card may with equal probability be any one of three honours; and accordingly the above written probability is to be multiplied by 3. If the turned-up card is not an honour then the probability that an assigned pair of honours is in the hands of the dealer or his partner, while the remaining two honours are in the hands of their adversaries, is 25/51⋅24/50⋅26/49⋅25/48; this probability is to be multiplied by six, as the assigned pair may be any of the six binary combinations formed by the four honours. Now the probability of the alternative first considered—the turned-up card being an honour—is 4/13; and the probability of the second alternative, 9/13. Accordingly the required probability is

${\frac {4}{13}}\cdot 3\cdot {\frac {25}{51}}\cdot {\frac {26}{50}}\cdot {\frac {25}{49}}+{\frac {9}{13}}\cdot 6\cdot {\frac {25}{51}}\cdot {\frac {24}{50}}\cdot {\frac {26}{49}}\cdot {\frac {25}{48}}={\frac {325}{833}}$.

34. The probability that each of the four players should have an honour may be calculated thus.^[8] If the card turned up is an honour then ipso facto the dealer has one honour and the probability that the remaining players have each an assigned one of the three remaining honours, is 13/51⋅13/50⋅13/49. Which probability is to be multiplied by 3!, as there are that number of ways in which the three cards may be assigned. If the card turned up is not an honour the probability that each player has an assigned honour is 13/51⋅13/50⋅13/49⋅12/48. Which probability is to be multiplied by 4!. Accordingly the required probability is

${\frac {4}{13}}\cdot 3!{\frac {13^{3}}{51\cdot 50\cdot 49}}+{\frac {9}{13}}\cdot 4!{\frac {12\cdot 13^{3}}{48\cdot 51\cdot 50\cdot 49}}={\frac {6\cdot 13^{3}}{51\cdot 50\cdot 49}}$.

(the chance not being affected by the character of the card turned up).

35. The probability of all the trumps being held by the dealer is

${\frac {12}{51}}\cdot {\frac {11}{50}}\cdots {\frac {2}{41}}\cdot {\frac {1}{40}}$ or ${\frac {12!39!}{52!}}$, which being calculated by means of tables for (logarithms of) factorials^[9] or directly,^[10] is 158,753,389,900.

36. There is a set of dominoes which goes from double blank to double nine (each domino presenting either a combination—which occurs only once—of two digits, or a repetition of the same digit). What is the probability that a domino drawn from the set will prove to be one assigned beforehand? The probability is the reciprocal of the number of dominoes: which is 10 × 9/2 (the number of combinations of different digits) + 10 (the number of doubles) = 55.

37. Choice and Chance.—When we leave the sphere of games of chance and frame questions relating to ordinary life there is a danger of assuming distributions of probability which are far from probable. For example, let this be the question. The House of Commons formerly consisting of 489 English members, 60 Scottish and 103 Irish, what was the probability that a committee of three members should represent the three nationalities? An assumption of indifference where it does not exist is involved in the answer that the required probability is the ratio of the number of favourable triplets, viz. 489 × 60 × 103 to the total number of triplets, viz. 652 × 651 × 650 × 3! A similar absence of selection is postulated by the ordinary treatment of a question like the following. There being s candidates

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1. Trans. Roy. Soc. (1895). See below, par. 165.
2. Todhunter, History . . . of Probability, and Bertrand, Calcul des probabilités, p. 9.
3. All three events cannot fail.
4. (c) occurring n times.
5. The reasoning may be illustrated by using the area of a circle to represent the frequency with which hearts fail, another (equal) circle for diamonds; for the case in which both hearts and diamonds fail the area common to the circles interlapping, and so on.
6. See Whitworth, Exercises in Choice and Chance, No. 502 (p. 125); referring to prop. xiv. of the same author's Choice and Chance.
7. Cf. Whitworth, Choice and Chance, question 143, p. 183, ed. 4.
8. Ibid.
9. There is such a table at the end of De Morgan's article in the Calculus of Probabilities in the Ency. Brit. “Pure Sciences,” vol. ii.
10. Cancelling factors common to the numerator and denominator.