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:: Gambler's Ruin Problem ::

Suppose two players P and Q are playing a gambling game until one of them has everything and the other is ruined. What is the probability that P will win everything? To answer this question, we can use any game where the probabilities of winning and losing are known.

For each play, P wagers £1 and receives £2 with probability and nothing with probability , i.e. gains or loses £1. In the real world, if Q were to be the casino, P would find that . Suppose the total amount of money between P and Q is £a and at some stage P has £x and Q has £(a-x). Let the probability of P winning everything be a function of x, f(x).

A game is played, if P won, he would have £(x+1), if he lost, he would have £(x-1) instead. Now if his fortune is £(x+1), his chances of winning everything is f(x+1), and f(x-1) if his fortune is £(x-1). Since the probability of winning a game is and losing the game , with a starting capital of x, where , we have the basic equation:

and in the limits of and ,

To solve the basic equation, we can try the solution

for some . This satisfies the basic equation, as shown below

Simplifying gives

which has the solution

Thus

Although both satisfy the basic equation, neither satisfies the criteria f(0) = 0 and f(a) = 1.Since 1 and satisfy the basic equation, so does this:

where A and B are constants. For f(0) = 0 and f(a) = 1,

Therefore for ,

This is the solution to the Gambler's Ruin problem, where the chances of winning everything is given by f(x), for given probability of winning a game , initial stake of and total stake of .

Extension to Fair Games

The solution above is valid only in situations where , for a fair game such as tossing a fair coin where , f(x) is reduced to

Mean Number of Rounds before Winning/Losing Everything

The mean or expected number of rounds before P wins or loses everything is given by (for )

For , this is simply

Written by Henry Tang.

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