Question

Gambler’s Ruin. A gambler, player A, plays a sequence of games against an opponent, player B. In each game, the probability of player A winning is p. If player A wins, he wins $1 which is paid by player B.

If he loses a hand with probability q = 1-p, he must pay $1 to player B.

The game ends either player B wins all the money from player A, and he is “ruined,” or when player A wins all the money from player B, and she is “ruined.”

If we say that the total pot is $N, we can model the game from player A’s perspective, that is, if player A has $k, then player B has $N-k.

Let y[k] = the probability of player A winning the game, given that he currently has a bankroll of $k. First-step Analysis: The game can be modeled as a random walk, where:

y[k] = p y[k+1] + q y[k-1] for 1 ≤ K ≤ N − 1

Boundary Conditions: y[0] = 0, player A is ruined, and B wins y[N] = 1, player B is ruined, and A wins

a) Find y[k], the probability of player A winning if p NOT EQUAL TO q?

b) Find y[k], the probability of player A winning if p = q = 0.5

c) Suppose both players start with the same amount of money, k = N-k. For p = 0.49, find y[k] for a. N=20 b. N=100 c. N=200

d) Comment on what the results imply from c) above.

Answer 1