Page:Encyclopædia Britannica, Ninth Edition, v. 19.djvu/795

Rh PROBABILITY 771 I. DETERMINATION OF THE PROBABILITIES OF COMPOUND EVENTS, WHEN THE PROBABILITIES OF THE SIMPLE EVENTS ON WHICH THEY DEPEND ARE KNOWN. 1. Under this head come a very large and diversified range of questions ; a very few of the most important are all that we can give. One great class relates to the fulfilment of given conditions in repeated trials as to the same event, knowing the probability of what will happen in each trial. 2. Lot there be an event which must turn out in one of two ways, W and B (as in drawing a ball from an urn containing white and black balls only) ; let the respective probabilities for each trial be p,q ; sothat j p + g f =l. Let two trials be made : the four possible cases which may arise are WW, WB, BW, BB. The probability of the first is p 1, of the second pq, of the third pq, of the fourth q 2. Thus the probability of a white and a black ball being drawn i an assigned order is pq ; but that of a white and a black in any order is 2/;g. Suppose now n trials to be made. The probability of W every time is p&quot; ; that of B once and W (n-l) times in an assigned order is p n l q, but if the order is indifferent it is np n ~ 1 q ; that of B occurring twice only is p n ~^q 2 if the order is given, but n(n 1 ) _Jpn-2qi i n anv order; and so on. We have then this a result : in the binomial expansion (p + q} n = 1 =^&quot; + np n - l q + - p n - 2 q 2 the terms in their order give the probabilities of the event W happening n times ; of W (n-1) times and B once ; of W (71 -2) times and B twice ; and so on, the sum of the whole giving 1, that is, certainty. 3. As an example, let A and B be two players whose respective chances of winning one game are p and q ; to find the probability of A winning in games before B wins ?i games, the play terminat ing when either of these events has occurred. The chance of A winning the first m games is p m . The chance of his winning in the first m + l games is mp m - 1 q. p = mp m q; for he must have won m - 1 games out of the first m, and then win the (tti + l)th; otherwise we should be including the first case. Again, the chance of A winning in the first m + 2 games is, in like (m + l)m (7)i + l)7)i manner, -^ p m - 1 q*p = p m q 2 ; and so on. Now the L 2 match must be decided at latest by the (m+n- l)th game ; for, if A fails to win ?)i games by that time, B must have won n. Hence the chance of A winning the match is ( m(m + l) m(m + l). . . (m + n -2), ,) p m 1 I + mq H ! ~ 2 - 1 - 1 n-l Thus, if A s skill be double that of B, the chance that A wins 112 131 four games before B wins two is ^-^ . That of B winning is . 24o 243 If A and B agree to leave off playing before the match is decided, the stakes ought clearly to be divided between them in proportion to their respective probabilities of winning, as given above, putting for m and n the numbers of games required to be won, at any given point of the match, by A and B respectively. This was one of the questions proposed to Pascal by the Chevalier de Mere in the year 1654. 4. In the expansion (1) it may be asked which combination of the events W, B is most likely to occur in the n trials. As the ratio of the 2d term to the 1st is n -. of the 3d to the 2d n - P 2 p and of the (r + l)th to the rth --, so long as this ratio continues to increase the terms will increase. The condition, therefore, for the rth term to be the greatest is n - r + 1 ^ p r J that is, r is the next integer above (71 + 1)7. We conclude that if r is the next integer below (n + l)q the (r + l)th term is the greatest that is, it is most likely that the event W occurs n-r times and B r times. If (n + l)q should be an integer (r), B is as likely to occur r as r+1 times ; and either is more probable than any other number. Thus, in twelve throws of a die, the ace is more likely to turn up twice than any other number; while in eleven throws it is as likely to turn up once only as twice. It is important to remark that, if the number of trials n be very large, we may treat qn and pi as whole numbers, and conclude that the event W is more likely to happen pn times and B qn times than in any other proportion. 5. Among the many questions which relate to the occurrence of different combinations in successive trials as to the same event, one is as to the chances for a succession, or run, of the same result several times. Let us consider the very simple case In n throws of a coin, what is the chance that head occurs (at least) twice running ? This will be an instance of the aid afforded by the calculus of finite differences in questions on probability. Let ?&amp;lt; r = the number of cases of r throws of a coin in which head turns up twice running, the whole number of cases being of course 2 r. Now if we consider the value of B + 3, it includes 2w n+2 , because the (?i + 3)th throw may turn up two ways ; but it includes also those cases when head turns up in the last two throws, tail in the preceding one. and no run of two heads occurs in the n preceding ones. The number of these cases is 2 n - u n. We have therefore the equation Un+Z = 2l( n+2 + 2&quot; - U n ..... (2). If E be an operator such that Ei r = w r +i, equation (2) is (E 3 -2E 2 + !) = 2&quot;; or, (E-l)(E 8 -E-l)w*-2&quot;; so that, if we put a, for the roots of the equation E 2 - E - 1 = 0, (3). since re =2&quot; is a particular solution of (2), A, B, C being three undetermined constants. Now in two throws there is one case where head turns up twice, and in three throws there are three cases ; hence we have u : = = 2 + A + Ba + C/3 we shall easily find = = and, remembering that o 2 = a + l, /3 2 = from these so that Now &amp;lt;- a a - n n+2 _ on+2 a-jB 1-V5 2 2 expanding by the binomial theorem and reducing, 71 + 2 l + : ^ . + l)7t(n-l)(7l-2), 2 (4). 5); dividing by the total number of cases 2 n, we have for the proba bility of head turning up at least twice running in 71 throws 6 . . . Another method of obtaining the same result is to consider the number of cases in which head never occurs twice running ; let u n be this number, then 2~ u n must be the number of cases when head occurs at least twice successively. Consider the value of + 2 ; if the last or (?i + 2)th throw be tail, w n + 2 includes all the cases (MB+J) of the n + 1 preceding throws which gave no succession of heads ; and if the last be head&quot; the last but one must be tail, and these two may be preceded by any one of the u n favourable cases for the first n throws. Consequently n +2 = U n+ i + U n , If a, j8, as before, are the roots of the quadratic E 3 - E - 1 = 0, this equation gives Here A and B are easily found from the conditions ^ = 2, 2 = viz., A B- a-/3 whence _ S- 5+&C

as in eq. (5). The probability that head never turns up twice running is found by dividing this by 2&quot;, the whole number of cases. This probability of course becomes smaller and smaller as the number of trials (n) is increased. 6. Let us consider the chance of a run of three heads or tails during n throws, that of a run of two heads or tails being 9 n 2 1 evidently =1 r, as there are but two cases out of the 2 n 2 n 2 n which are alternately head and tail. Let 11 r be the number of cases, during r throws, which give at least one succession of three heads or three tails. Consider the value of u H+3 ; it includes 2u,,+ s, as the last throw may be head or tail ; but, besides these, every case of the first n throws which contains no run of three gives rise to one new case of the n + 3 having a run of three ; thus, if the ?th throw be head, the last four may be HTTT, or THHH if the Tith be tail. Hence