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盲弓手第三箭

The Blind Archer

专题
Probability / 概率
难度
L4

题目详情

一位稳定且技术一致的盲弓手朝圆靶中心射了两箭。其每次射击偏差随机，但分布相同。

他被告知“第一箭比第二箭更靠近中心”。他又射了第三箭。

问：第三箭成为三箭中最靠近中心的概率是多少？

提示：对称性/枚举。

A very sharp, consistently skillful blind archer aimed for the center of a circular board and shot 2 arrows. He is expected to hit the aim, but doesn't hit it for sure. The archer is told that his first shot is better than second. He tried one more shot. What is the probability that this 3rd shot is the best shot among 3? (ie, Probability that 3rd arrow lands closer to center than his first two shots?)

Hint

Enumeration; Symmetry between random variables.

解析

答案是 13\frac{1}{3}。

设三箭到中心距离为 x1,x2,x3x_1,x_2,x_3。在不带条件下，x1,x2,x3x_1,x_2,x_3 的 6 种大小排序等可能。

已知 x1<x2x_1<x_2，剩下等可能的三种情况是：

  • x1<x2<x3x_1<x_2<x_3
  • x1<x3<x2x_1<x_3<x_2
  • x3<x1<x2x_3<x_1<x_2

其中只有最后一种使第三箭最好，因此概率为 1/31/3。

Original Explanation

1/3

Solution

Suppose x1, x2 and x3 are the distances of the arrows from center. As the archer is consistent, we can use symmetry, i.e., there are six equally likely cases: x1<x2<x3, (or 5 others, which are its permutations). Since archer is told that x1<x2, we are left with following equally likely cases: x1<x2<x3 , x2<x1<x3 , x1<x3<x2 Among these three, one is favorable. Hence The probability of last shot being best is 1/3.

Notice that if there were (N-1) arrows, with first being better than rest, and then he shoots Nth arrow. The probability that Nth shot is best is 1/N.