TITLE: Two cowboys A and B decide to solve a dispute with a duel.
QUESTION [7 upvotes]: Two cowboys $A$ and $B$ decide to solve a dispute with a duel.
Cowboy $A$ hits his target $\frac13$ of the time.
Cowboy $B$ hits the target $\frac23$ of the time. 
It is decided that $A$ will take the first shot, cowboy $B$ will take the second shot (if still alive).
This will continue until there is only one left alive. Also, a cowboy can not shoot two times in a row. What are cowboy $A$ chances of winning the duel?
My logic is to simulate the cases where $A$ wins:
I Case:  $A$ wins: $\frac13$
II Case: $A$ loses, $B$ loses, $A$ wins: $\frac23\cdot\frac13\cdot\frac13$
...
But I don't know what to do from here. Any help?

REPLY [8 votes]: Let $p$ be the probability that A wins. With probability ${1\over3}$ he wins on his first shot. With probability ${2\over3}$ he misses. When B hits in his first shot, A has lost. When B misses, A again has a chance of $p$.  It follows that
$$p={1\over3}+{2\over3}\cdot{1\over3}\cdot p\ .$$
This implies $p={3\over7}$.