# Probability, poor and good shooter

Imagine you have a duel with two opponents, Person $B$ is a poor shooter with probability of shooting his opponent equaling $1/3$, while person $A$ is a good shooter with probability of shooting his adversary equaling $2/3$. The rules of duel are the following: first shoots $B$, if $B$ misses, then person $A$ shoots. If $A$ misses then again $B$ shoots, and so on, until one shooter stands alive. What is the probability that shooter $B$ wins the duel?

After just $1$ round, $B$ wins with probability $\frac{1}{3}$ and $A$ wins with probability $\frac{2}{3}\times\frac{2}{3}$.
So chances are divided $3$ for $B$ and $4$ for $A$.
$B$ wins $\frac{3}{7}$ times.
Let $P$ be the probability that $B$ wins the whole duel, then $P = \frac13 + \frac23(1-\frac23)P$. The $\frac13$ is the probability that B wins the current duel, $\frac23$ is the probability that $B$ misses and the $(1-\frac23)$ is the probability that A misses. If A misses we are in the same situation in which we started.
So $P = \frac13 + \frac23(1-\frac23)P = \frac13 + \frac29 P \implies P = \frac37$.