Player A has 2 coins, and player B has 3 coins. They flip a biased coin and the winner pays 1 coin to the loser. The chance of A winning is p. They stop playing when one of the players run out of coins.
(a) What is the probability of A having all the coins at the end? (b) What is the expected number of coin flips?
I am having trouble finding the probability that A or B wins. Any help would be much appreciated.
Let $x_k$ denote the probability player $A$ wins with $k$ coins in his wallet. We want to compute $x_2$, and note that $x_5 = 1$ and $x_0 = 0$. By conditioning, it holds $$x_k = px_{k+1} + (1-p)x_{k-1}.$$ Then you will get the following system of linear equations:
$$x_1 = px_2$$ $$x_2 = px_3 + (1-p)x_1$$ $$x_3 = px_4 + (1-p)x_2$$ $$x_4 = p + (1-p)x_3.$$ Solving for the unknowns $\{x_1,x_2,x_3,x_4\}$ will tell you the probabilities of winning. For reference, this is related to a classic Markov chain problem called "gambler's ruin."