俄罗斯轮盘 II

Russian Roulette II

专题: Probability / 概率
难度: L4
来源: QuantQuestion

题目详情

你和朋友玩俄罗斯轮盘：6 膛左轮装 1 发子弹。每次轮到某人扣扳机之前，都会重新转动弹巢使其随机。

两人轮流扣扳机，直到有人中弹输掉。

若朋友先手，你获胜的概率是多少？

You and your friend decide to play a game of Russian Roulette. The gun is a six-chambered revolver and it's loaded with one bullet. To start each round, the cylinder is spun to randomize the order of the chambers. The two of you take turns pulling the trigger until the person who fires the gun loses. Given that the cylinder is re-spun after each turn, what is the probability that you win if your friend goes first?

解析

设 $p$ 为先手（朋友）最终输掉的概率。

第一次扣扳机：先手以 $1/6$ 概率立刻输；以 $5/6$ 概率存活并把局面交给后手，而后手成为“先手的对手”，其输掉概率为 $1-p$ 。

因此

p=\frac{1}{6}+\frac{5}{6}(1-p) \Rightarrow p=\frac{6}{11}.

朋友输即你赢，所以你获胜概率为 $6/11$ 。

Original Explanation

We'll have $p$ represent the probability that your friend, the first player, loses. Then $1-p$ will be the probability that you, the second player, lose. We will condition the probability $p$ on the first trigger pull. The first player loses with certainty $\frac{1}{6}$ of the time. Otherwise, he essentially becomes the second player in the game with a conditional probability $1-p$ of losing. Thus,

p = \frac{1}{6} \times 1 + \frac{5}{6} \times (1-p) \Rightarrow p = \frac{6}{11}