MathProblem 33 Russian roulette problem

Assuming both players take turns what is the probability the player who goes first will lose at Russian roulette using a gun with six chambers?

Solution

两个人轮流玩 \(6\) 发左轮手枪，问先手输的概率是多大。

设先手赢的概率为 \(p\), 后手赢的概率为 \(q\): \(p+q=1\)

对于概率 \(q\):

\[q=P_1(\text{survive after first shoot})\cdot P_2(\text{win after survive}) \]

显然 \(P_1=5/6\)，对于 \(P_2\)，当后手生存以后，此时后手转变为先手，因此：

\[q=\frac{5}{6}p \]

带入 \(p+q=1\)，求解即可。