The simplified version of a (newspaper column) problem
goes like this:

A starts with 2 dollars and B with 1 dollar. They play
a game by tossing a fair coin. If it comes up heads,
A gives B one dollar. If it comes up tails, B gives A
one dollar.The game ends when either one runs out
of coins. What are the chances that A will win?

It is not difficult to draw out the tree in this case.
To win the game

1. A should "win" the first toss
2. Lose the first, and win the next two
3. ....

Essentially, to win his path is

   (LW)^n W, with n = 0, 1, 2, .....

where 'L' stands for losing and 'W' stands for
winning a toss.

The geometric sum gives the probablity as 2/3.

What I am after is the more complicated case of the
starting situation with A having 3 dollars and B with 2,
which is the problem in the newspaper.
Here the game does not terminate as easily as in the
previous case as there are many paths the "loop back"
to the starting configuration, unilke the "only way" in the
previous case. As the numbers increase for the
starting configuration, a player can recover even after
going into a prolonged losing streak -- something not
possible in the example.

What I am wondering is whether there is some sort of
recursive relation that can be written down to get the
probability of one of the players winning the game.

One wild guess is that in the starting state was (m, n),
the probability that A will win is

 m/(m+n)

but I doubt life is that simple!

thanks
mt

On Sat, 10 Nov 2012 19:35:22 -0000, "Mike Terry"

Thanks for the hint. Had never heard of "The Gambler's
Ruin" in statistics or Martin Gardner or other readings.

I guess that the idea of using a recursive approach is
the usual one, but I was badly off in the way I
wanted to use it. I was tryig to relate the probability
of wining from a certian position to itself after some
moves that brought back the intial state. But, thiss
leads to problems as there are too many ways that
it can come about. The "state transition" approach
is neat -- I was to fixated with my approach to
think of it.

Well I don't know how different from state transition it appears, but many such problems look amenable to the following approach:

Let W be the total wealth - here "5".  Then if A has x, B must have W-x.  
Probability of A winning at x is:
p(x) = 1/2 p(x+1) + 1/2 p(x-1)
as those are the only two outcomes possible at the next toss.  So clearly we always have the probability as an average of nearby probabilities.  The extreme ends (x = 0 or W) clearly can be assigned 0 and 1 probabilities.  Now solving should be easy for p(x) or p(3) specifically...

HTH.