已知第二镖更远，第三镖超过第一镖概率

Dart Game

专题: Probability / 概率
难度: L4
来源: QuantQuestion

题目详情

Jason 连续掷两镖瞄准靶心，且已知第二镖落点距离靶心比第一镖更远。

如果他再掷第三镖（同样水平稳定），第三镖落点距离靶心比第一镖更远的概率是多少？

Jason throws two darts aiming at the center of a dartboard. The second dart lands farther from the center than the first. If Jason throws a third dart, also aiming at the center, what is the probability that the third throw ends up farther from the center than the first throw? Assume constant skillfulness.

解析

设三次距离 $D_1,D_2,D_3$ 为独立同分布且连续。

在 6 种大小排序中等可能。条件 $D_2>D_1$ 使得只剩 3 种等可能排序：

$D_1<D_2<D_3$
$D_1<D_3<D_2$
$D_3<D_1<D_2$

其中前两种满足 $D_3>D_1$ ，故概率为 $\frac{2}{3}$ 。

Original Explanation

Method 1 (Enumerating Orders)
Name the three throws A, B, C in order of distance from center. There are 6 equally likely permutations if all are distinct. Given that the second throw (call it “B”) is farther than the first (“A”), that condition excludes half of these. Among those allowed, in 2 out of 3, the third throw (“C”) is farther than “A.” Hence the probability is $\frac{2}{3}.$
Method 2 (Symmetry Argument)
Equivalently, it is the probability that the third dart is not the best throw among all three. By an order-statistic or symmetry argument, that probability is $\frac{2}{3}$ .