Winning the GameDate: 09/21/2000 at 14:29:49 From: Ryan Lane Subject: Mathematical Expectation In an equitable game of coin tossing, I have x dollars and my friend has y dollars. I win a dollar if I choose correctly after each flip, but my friend loses a dollar, and vice versa. What is the probabitity that my friend will win all of my x dollars before I can win his y dollars, if each of our mathematical expectations is zero? I am stuck because I don't know if an integral is appropriate for the question. Thank you for your time. Date: 09/22/2000 at 08:24:02 From: Doctor Anthony Subject: Re: Mathematical Expectation It is agreed that the series of games is won by A if he wins x dollars more than B before B wins y dollars more than A, and is won by B in the contrary event. In effect, A has y dollars at the start and B has x dollars at the start. Let u(n) be the probability of A winning the series considered at the stage when he has already won n games more than B, where -y < n < x. This is a valid definition because the conditions of winning the series are concerned only with the difference between the number of games won by A and B respectively and not with the absolute numbers themselves. To determine a forward difference equation for u(n), we proceed as follows. p = probability of A winning a game, q = probability of B winning a game where p + q = 1 u(n) = p.u(n+1) + q.u(n-1) That is, there is probability p of moving to u(n+1), and probability q of moving to u(n-1) So we get the difference equation p.u(n+1) - u(n) + q.u(n-1) = 0 with auxiliary equation p.m^2 - m + q = 0 and so (m-1)(pm - q) = 0 therefore m = 1 or m = q/p and u(n) = A + B.(q/p)^n If p = q, the m equation has a double root m = 1 and we get u(n) = A + B.n Now u(x) = 1 and u(-y) = 0 and so 1 = A + B.(q/p)^x and 0 = A + B.(q/p)^(-y) and these simplify to A = 1/[1 - (q/p)^(x+y)] B = -(q/p)^y/[1 - (q/p)]^(x+y)] 1 - (q/p)^(n+y) and so u(n) = --------------- if p <> q 1 - (q/p)^(x+y) If p=q the equations for A and B are 1 = A + Bx 0 = A - By giving A = y/(x+y), B = 1/(x+y) and u(n) = (y+n)/(x+y) At the start these equations give 1 - (q/p)^y u(0) = -------------- for p <> q 1 - (q/p)^(x+y) and u(0) = y/(x+y) if p=q So the initial probability that A will win x dollars from B before B wins y dollars from A is y/(x+y) when p = q. This is to be expected, since A starts with y dollars and B starts with x dollars, and the game ends when either has exhausted his capital. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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