What's the probability that A wins finally Suppose A has \$2 and B has \$3. They play a game, each game gives the winner \$1 from other. A has a probability $\frac{3}{5}$ to win each game. They play this game until one of them is bankrupt. What's the probability that A wins finally?
 A: Hint: denote $X_t$ as the cash amount of A, then $\left(\frac{1-3/5}{3/5}\right)^{X_t}$ is a martingale.
A: How to do it depends on the tools you have. We use only basic techniques. 
Let $p_1$ be the probability that A (ultimately) wins the game if she currently has $1$ dollar. 
Let $p_2$ be the probability she ultimately wins if she has $2$ dollars.  Define $p_3$ and $p_4$ analogously.  We only want to know $p_2$, but the others will be helpful.
To introduce fancier language, $p_i$ is the probability she ultimately wins given that she is in State $i$.
We have $p_1=\frac{3}{5}p_2$. For with probability $\frac{2}{5}$ she will lose the game, and be broke. With probability $\frac{3}{5}$ she will win the game, will have $2$ dollars,  and then her probability of ultimately winning is by definition $p_2$.
Similarly, $p_2=\frac{2}{5}p_1+\frac{3}{5}p_3$.  
Write down analogous equations for $p_3$ and $p_4$, carefully.
You will end up with $4$ (easy) linear equations in $4$ unknowns. Solve for $p_2$. 
