队伍中第几位最可能拿到“生日重复奖”

Birthday Line

专题: Probability / 概率
难度: L4
来源: QuantQuestion

题目详情

电影院经理宣布：队伍里第一个出现“生日与前面某人重复”的人将获得免费票。

你可以选择站在队伍的任意位置 $n$ 。

假设生日在 365 天均匀分布，问选哪个位置能最大化你成为获胜者的概率？

A movie theater manager announces: the first person in line whose birthday matches someone who already bought a ticket in that line wins a free ticket. You can choose any position $n$ in the line. Assume birthdays are uniformly distributed over 365 days. Which position maximizes your chance of being the winner?

解析

结论：最佳位置约为 第 20 位。

第 $n$ 位获胜需要：前 $n-1$ 位生日全不同，且第 $n$ 位生日与前面某人重复。

P(n)=\frac{365\cdot 364\cdots (365-(n-2))}{365^{n-1}}\cdot\frac{n-1}{365}.

分析 $P(n)$ 的增减可得最大值在 $n=20$ 附近。

Original Explanation

Define $P(n)$ as the probability the $n$ -th person is the first to match a previous birthday. Then $P(n) = \biggl(\frac{365 \times 364 \times \cdots \times (365 - (n-2))}{365^{\,n-1}}\biggr)\,\times \frac{n-1}{365}.$ Analyzing the increase/decrease behavior around $n$ shows that $n=20$ is the position that maximizes this probability.