Winning the Game

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.

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/