俄罗斯轮盘 I

Russian Roulette I

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 发左轮手枪装 1 发子弹。开始时转动弹巢一次使其随机化，之后不再重新转动。

两人轮流扣扳机，直到某人开到子弹输掉。

若朋友先开枪，你获胜的概率是多少？

You and your friend are bored one day and decide to play Russian Roulette. The six-chambered revolver is loaded with one bullet. To start the game, 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. Considering the cylinder is not re-spun after each turn, what is the probability that you win if your friend goes first?

解析

初始转动后子弹位置等概率落在 1..6 任一膛。

朋友先开枪，你会在子弹位于第 2、4、6 膛时输；在第 1、3、5 膛时赢。

因此你赢的概率为 $3/6=1/2$ 。

Original Explanation

Once the cylinder is initially spun, the position of the bullet is fixed. As your friend is going first, you would win if the bullet is in chambers 1, 3, or 5, and lose if the bullet is in chambers 2, 4, or 6. The probability of winning is $\frac{1}{2}$ .