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### Winning the Game

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Date: 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.

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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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Associated Topics:
High School Probability

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