Russian roulette with a re-spin, going second

Two players take turns with a six-chamber revolver holding one bullet. Before every pull of the trigger the cylinder is spun again. Whoever fires the bullet loses. If your opponent pulls first, what is your probability of winning?

$\tfrac{6}{11}$. Let $p$ be the chance the first shooter loses. They lose immediately with probability $\tfrac16$, and otherwise become the second shooter: $p=\tfrac16+\tfrac56(1-p)$, so $p=\tfrac{6}{11}$, which is your chance of winning.