100 个点球：最终 66 球的概率

100 penalty kicks

专题: Probability / 概率
难度: L4
来源: QuantQuestion

题目详情

共有 100 次点球。

第 1 次必进（Goal），第 2 次必丢（Miss）。
对于之后的每一次 $i\ge3$ ：进球概率为

\frac{\text{到目前为止的进球数}}{\text{到目前为止的尝试次数}}.

问：最终 100 次里恰好进 66 球的概率是多少？

We have a sequence of 100 kicks. Kick #1 is a goal, kick #2 is a miss. Then for each subsequent kick $i\ge3$ , the probability of scoring is $\frac{\text{number goals so far}}{\text{number attempts so far}}$ . We want the probability that the final total goals (out of 100) is exactly 66.

解析

这类过程可视作 Polya urn / 强化过程，也可用鞅观点分析。

在前两次固定为 G、M 后，你已有 1 球进、1 球丢。之后 98 次的分布具有对称性：在“以当前命中率作为下一次命中概率”的规则下，给定首两次结果后，后续序列在某种意义上等概率。

要最终总进 66 球，由于前两次已进 1 球，因此接下来 98 次需要再进 65 球。于是概率可写为

\frac{\binom{98}{65}}{\binom{100}{66}}.

Original Explanation

This is a Polya’s urn type problem or a “martingale” approach. In fact, the final distribution (given the first 2 outcomes fixed) ends up with each sequence equally likely among sequences that begin G, M. More direct approach: after 2 kicks, you have 1 goal, 1 miss. For the next 98 kicks, each arrangement of 65 additional goals + 33 additional misses is equally likely. You just want total 66 goals. But you already have 1 goal in the first 2. So you need 65 more goals in the next 98. The probability is: $\frac{\binom{98}{65}}{\binom{100}{66}},$ since all ways of having 66 total out of 100 are equally likely if the fraction-based probability is used (this is a known property of the Polya/Eggen distribution). Another approach is the Beta-Binomial formula.