Page:EB1911 - Volume 22.djvu/394

Rh “The difficulty consists in the enumeration of the cases,” as Lagrange says. Sometimes summation is the only mathematical operation employed; but very commonly it is necessary to apply the theory of permutations and combinations involving multiplication.

22. Fundamental Theorem.—One of the simplest problems of this sort is one of the most important. Given a mélange of things consisting of two species, if n things are taken at random what is the probability that s out of these n things will be of a certain species? For example, the mélange might be a well-shuffled pack of cards, and the species black and red; the quaesitum, what is the probability that if n cards are dealt, s of them will be black? There are two varieties of the problem: either after each card is dealt it is returned to the pack, which is reshuffled, or all the n cards are dealt (as in ordinary games of cards) without replacement. The first variety of the problem deserves its place as being not only the simpler, but also the more important, of the two.

23. At the first deal there are 26 cases favourable to black, 26 to red. When two deals have been made (in the manner prescribed), out of 522 cases formed by combinations between a card turned up at the first deal and a card turned up at the second, 26 × 26 cases are combinations of two blacks, 26 × 26 are combinations of two reds, and the remainder 2(26 × 26) are made up of combinations between one black and one red; 26 × 26 cases of black at the first deal and red at the second, and 26 × 26 cases of red at the first and black at the second deal. The number of cases favourable to each alternative is evidently given by the several terms in the expansion of (26 + 26)2. The correspond in probabilities are given by dividing each term by the total number of cases, viz. 522. Similarly, when we go on to a third deal, the respective probabilities of the three possible cases, three blacks, two blacks and one red, two reds and one black, three blacks, are given by the successive terms in the binomial expansion of (26 + 26)3, and so on. The reasoning is quite general. Thus for the event which consists of dealing either clubs or spades (black) we might substitute an event of which the probability at a single trial is not ½, e.g. dealing hearts. Generally, if p and 1 − p are the respective probabilities of the event occurring or not occurring at a single trial, the respective probabilities that in n trials the event will occur n times, n − 1 times. . . twice, once or not at all, are given by the successive terms in the expansion of [p+ (1 − p)]n; of which expansion the general term is $$\frac{n!}{s!(n-s)!}p^s{(1-p)}^{n-s}$$.

24. The probability may also be calculated as follows. Taking for example the case in which the event consists of dealing hearts; consider any particular arrangement of the n cards, of which s are hearts, e.g. the arrangement in which the s cards first dealt are hearts and the following n − s all belong to other suits. The probability of the first s cards being all hearts is $$(\frac{1}{4})^s$$; the probability that none of the last (n − s) cards are hearts is $$(\frac{3}{4})^{n-s}$$. Hence the probability of that particular arrangement occurring is $$(\frac{1}{4})^s(\frac{3}{4})^{n-s}$$. But this arrangement is but one of many, e.g. that in which the s hearts are the last dealt, which are equally likely to occur. There are as many different arrangements of this type as there are combinations of n things taken together s times, that is $$\frac{n!}{s!(n-s)!}$$ The probability thus calculated agrees with the preceding result.

25. It follows from the law of expansion for [p+ (1 − p)]n that as n is increased, the value of the fractions which form the terms at either extremity diminishes. When n becomes very large, the terms which are in the neighbourhood of the greatest term of the expansion overbalance the sum total of the remaining terms. Thus in the example above given, if we go on and on dealing cards (with replacement) the ratio of the red cards dealt to all the cards dealt tends to become more and more nearly approximate to the limit ½. These statements are comprised in the theorem known as James Bernoulli's. Stated in its simplest form—that “in the long run all events will tend to occur with a relative frequency proportional to their objective probabilities” —this theorem has been regarded as tautological or circular. Yet the proofs of the theorem which have been given by great mathematicians may deserve attention as at least showing the consistency of first principles. Moreover, as usually stated, James Bernoulli's imports something more than the first axiom of probabilities.

26. The generalization of the Binomial Theorem which is called

the “Multinomial Theorem” gives the rule when there are more than two alternatives at each trial. For instance, if there are three alternatives, hearts, diamonds or a card belonging to a black suit, the probability that if n cards are dealt there will occur s hearts, t diamonds, and n − s − t cards which are either clubs or spades is

27. Applications of Fundamental Theorem.—The peculiar interest of the problem which is here placed first is that its solution represents a law of almost universal application: the law assigning the frequency with which different values assumed by a quantity, like most of the quantities with which statistics has to do, depends upon several independent agencies. It is remarkable that the problem in probabilities which historically was almost the first belongs to the kind which is first in interest. Of this character is a question which occupied Galileo and before him Cardan, and an even earlier writer: what are the chances that, when two or three dice are thrown, the sum of the points or pips turned up should amount to a certain number? A particular case of this problem is presented by the old game of “passedix”: what is the probability that if three dice are thrown the sum of the pips should exceed ten? The answer is obtained by considering the number of combinations that are favourable to each of the different alternatives, 18 pips, 17, 16. . . . 11 pips, which make up the event in question. Thus out of the total of 216 (63) combinations, one is favourable to 18, three to 17, and so on. There are twenty-five chances, as we may call the permutations, in favour of 12, twenty-seven in favour of 11. The sum of all these being 108, we have for the event in question 108/216, an even chance. More generally it may be inquired: what is the probability that, if n dice are thrown, the number of points turned up will be exactly s? By an extension of the reasoning which was employed in the first problem it is seen that the required probability is that of which the index is s in the expansion of the expression

The calculation may be simplified by writing this expression in the form

The successive terms of the expansion give the respective probabilities that the number in question should be n, n + 1. . . 6n comprising all the possible numbers among which s is presumably included (otherwise the answer is zero). Of course we are not limited to six alternatives; instead of a die we may have a teetotum with any number of sides. The series expressing the probabilities of the different sums can be written out in general terms, as Laplace and others have done; but it seems to be of less interest than the approximate formula which will be given later.

28. Variant of the Fundamental Theorem.—The second variety of our first problem may next be considered. Suppose that after each trial the card dealt (ball drawn, &c.) is not replaced in statu quo ante. For instance, if r cards are dealt in the ordinary way from a shuffled pack, what is the probability that s of them will be hearts (s < 13)? Consider any particular arrangement of the r cards, whereof s are hearts, e.g. that in which the s cards first dealt are all hearts, the remaining r − s belonging to other suits. The probability of the first card being a heart is ; the probability that, the first having been a heart, the second should be a heart is (since a heart having been removed there are now 12 favourable cases out of a total of 51 cases). And so on. Likewise the probability of the (s + 1)th card being not a heart, all the preceding s having been hearts, is 39/(52 − s), the probability of the (s + 2)th card being not a heart is similarly reckoned. And thus the probability of the particular arrangement considered is found to be

Now consider any other arrangement of the r cards, e.g. t of the s hearts to occur first and the remaining s − t last. The denominator in the above expression will remain the same; and in the numerator only the order of the factors will be altered. The probability of the second arrangement is therefore the same as that of the first; and the probability that some one or other of the arrangements will occur is given by multiplying the probability of any one arrangement and the number of different arrangements, which, as in the simpler case of the problem, is the same as the number of combinations formed by r things taken together s times, that is r!/s!(r − s)!. The formula thus obtained may be generalized by substituting n for