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专题
Probability / 概率
难度
L7

题目详情

你在练习足球点球，共踢 100 次。已知第 1 球踢进了，第 2 球踢丢了。从第 3 球开始，每一脚踢进的概率等于此前已踢进球数占尝试总数的比例。比如，若前 30 次尝试中进了 17 球，则第 31 球踢进的概率为 1730\frac{17}{30}。包含前两脚在内，100 次尝试后你恰好进 66 球的概率是多少？

You're practicing soccer by taking 100 penalty kicks. Assume that you have made the first goal but missed the second. For each of the following kicks, the probability that you score is the fraction of goals you've made thus far. For example, if you made 17 goals out of the first 30 attempts, then the probability that you make the 31st goal is 1730\frac{17}{30}. After 100 attempts, including the first two, what is the probability that you score exactly 66 penalty kicks?

解析

设 N=100N=100，KK 表示踢完 NN 脚后的总进球数。题目已知前两脚结果分别为：第 1 脚进球，第 2 脚未进。因此前两脚后我们有 S2=1S_2=1 次成功、F2=1F_2=1 次失败。对每个后续时刻 t≥3t\geq3，进球概率为：

Pr(score at t | history)=St−1t−1\text{Pr(score at t | history)} = \frac{S_{t-1}}{t-1}

未进球的概率则为 t−1−St−1t−1\frac{t-1-{S_{t-1}}}{t-1}。

固定任意一个由第 3,4,…,N3,4,\ldots,N 脚组成的具体结果序列，并且它最终使总进球数 K=kK=k（包含第 1 脚的进球）。这意味着在剩余的 N−2N-2 脚中，恰好多了 k−1k-1 个进球和 N−k−1N-k-1 个未进球。

来计算这个特定序列的概率。分母会贡献一个因子 ∏m=2N−1m=(N−1)!∏_{m=2}^{N-1}m = (N-1)!，因为对每个 t=3t=3 到 NN，我们都要除以 t−1t-1，而这些 (t−1)(t-1) 的乘积就是 2∗3∗⋯∗(N−1)=(N−1)!2 * 3 * \cdots * (N-1) = (N-1)!。

分子则由每次成功或失败前对应的当前计数累乘而成。随着成功一次次出现，成功事件（除去第一脚已知成功）对应的分子恰好是整数

1,2,...,k−11, 2, ..., k-1

因此它们的乘积为 (k−1)!(k-1)!。同理，失败事件（除去第二脚已知失败）对应的分子是：

1,2,...,N−k−11, 2, ..., N-k -1

因此它们的乘积为 (N−k−1)!(N-k-1)!。所以，在前两脚结果已固定的情况下，该特定有序序列的概率为：

(k−1)!(N−k−1)!(N−1)!\frac{(k-1)!(N-k-1)!}{(N-1)!}

那么，剩余 N−2N-2 脚中恰有 k−1k-1 次成功的不同有序序列一共有多少个？答案正是

(N−2k−1)\begin{pmatrix} N-2\\ k-1 \end{pmatrix}

即从这 N−2N-2 个位置中选出 k−1k-1 个作为成功的位置。将单个序列的概率乘上这样的序列个数，得到

Pr⁡(K=k)=(N−2k−1)⋅(k−1)! (N−k−1)!(N−1)!=(N−2)!(N−1)!=1N−1.\Pr(K = k) =\binom{N-2}{k-1}\cdot\frac{(k-1)!\,(N-k-1)!}{(N-1)!} =\frac{(N-2)!}{(N-1)!} =\frac{1}{N-1}.

因此，KK 在整数 k=1,2,…,N−1k=1,2,\ldots,N-1 上服从均匀分布（在第 1 脚进、第 2 脚失的条件下，这正是所有可能的总进球数）。对于本题 N=100N=100，有：

Pr(K=66)=1100−1=199\text{Pr(K=66)} = \frac{1}{100-1} = \boxed{\frac{1}{99}}

Original Explanation

Let N=100N=100. Let KK be the total number of goals after NN kicks. We are told the outcomes of the first two kicks: goal on kick 1 and miss on kick 2. So after the first two kicks we have S2=1S_2=1 success and F2=1F_2 =1 failure. For each subsequent kick t≥3t\geq3 the probability of scoring is:

Pr(score at t | history)=St−1t−1\text{Pr(score at t | history)} = \frac{S_{t-1}}{t-1}

and the probability of missing is t−1−St−1t−1\frac{t-1-{S_{t-1}}}{t-1}

Fix any particular sequence of outcomes for kicks 3,4,....,N3, 4, ...., N that results in a total of K=kK=k goals (including the first goal). This means among the N−2N-2 remaining kicks there are exactly k−1k-1 additional goals and N−k−1N-k-1 additional misses.

Compute the probability of that specific sequence. The denominator contributes a factor of ∏m=2N−1m=(N−1)!∏_{m=2}^{N-1}m = (N-1)! (because for each tt from 33 to NN we divide by t−1t-1; the product of those (t−1) is 2∗3...(N−1)=(N−1)!(t-1) \ \text{is} \ 2 * 3 ... (N-1) = (N-1)!).

The numerator collects factors given by the current counts just before each success or failure. As successes occur one by one, the numerators for the success events (after the first) are exactly the integers

1,2,...,k−11, 2, ..., k-1

so their product is (k−1)!(k-1)!. Likewise the numerators for the failure events (after the second) are:

1,2,...,N−k−11, 2, ..., N-k -1

so their product is (N−k−1)!(N-k-1)!. Therefore the probability of that particular ordered sequence (with the first two outcomes fixed) is:

(k−1)!(N−k−1)!(N−1)!\frac{(k-1)!(N-k-1)!}{(N-1)!}

How many different ordered sequences of the remaining N−2N−2 kicks give exactly k−1k−1 successes? Exactly

(N−2k−1)\begin{pmatrix} N-2\\ k-1 \end{pmatrix}

choices of positions for the k−1k-1 successes. Multiplying the per-sequence probability by the number of such sequences gives

Pr⁡(K=k)=(N−2k−1)⋅(k−1)! (N−k−1)!(N−1)!=(N−2)!(N−1)!=1N−1.\Pr(K = k) =\binom{N-2}{k-1}\cdot\frac{(k-1)!\,(N-k-1)!}{(N-1)!} =\frac{(N-2)!}{(N-1)!} =\frac{1}{N-1}.

Thus KK is uniform on the integers k=1,2,...,N−1k=1, 2,..., N-1 (these are exactly the possible totals given the first is a goal and the second a miss). For our case N=100N=100 we get:

Pr(K=66)=1100−1=199\text{Pr(K=66)} = \frac{1}{100-1} = \boxed{\frac{1}{99}}