Someone wants to play the Gambler’s Ruin game with you, where you pay them $1$ dollar if the coin comes up heads and they pay you $1$ if it comes up tails. You discover that a biased coin will be used that has probability $2/3$ of coming up heads. Even though the other player seems to have the advantage, you get to start with $2$ dollars and they will start with $1$. Whoever gets all $3$ dollars first will win the game. Do you still have an advantage, even though you use a biased coin?
Find out by computing $p_1$, where $p_i$ is the probability that the other player wins the game starting with i dollars, for $0\leq i\leq3$. Give an exact value for $p_1$ as a simplified fraction.
How do I go about solving this problem?
I understand the general formula for conditional probability, and in this case I would start with $$P(E_i|H)P(H) + P(E_i|T)P(T)$$
But since "i" doesn't seem to have an upper bound, I don't really know how to compute $p_i$, let alone $p_1$.
Any help is appreciated. Thanks!
Given the instructions, do the following:
Let $p_i$ be the probability you lose if you are playing this game and currently you have $3-i$ dollars and they have $i$. Call this "state" $i$.
Then note that $p_0=0$: since they have $0$, they have already lost the game.
The interesting cases are the middle ones, where you can write a recursive relationship. Each round of the game the opponent either wins $1$ (with probability 2/3) or loses $1$ (with probability 1/3). So, after one round you move from state $i$ to either $i+1$ or $i-1$.
By definition of $p_i$ (the winning probability, given that we started at state $i$) we therefore have $$p_1=\frac{2}{3}p_2+\frac{1}{3}p_0$$
You can solve for all the $p_i$s from here.