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网球随机配对：无女女对战的概率

A Tournament Problem

专题
Strategy / 策略
难度
L4

题目详情

Ten players participate in the first round of a tennis tournament: 2 females and 8 males. Five single matches are fixed at random by successively drawing, without replacement, the names of all 10 players from an urn: the player drawn first plays against the one whose name comes up second, the third against the fourth, etc.

a. What is the probability that there will not be a single match involving two female players? Is this probability smaller, equal to, or larger than the corresponding probability with 20 females and 80 males?

b. Try to answer the general case in which there are 2n players, of whom 2<k<n2 < k < n are female. What is the probability p(k,n)p(k, n) that among the n matches there will not be a single one involving two female players?

解析

(a) 10 人中 2 女 8 男。等价于：第一位女性的对手必须是男性。随机从其余 9 人中选对手，男性有 8 个，所以

P(无女女对战)=89.\boxed{\mathbb{P}(\text{无女女对战})=\frac{8}{9}}.

当人数变大且女性比例固定时，期望出现的“女女对”数随 nn 增大而增大，因此“无女女对战”的概率会更小；所以 2 女 8 男的概率 更大。

(b) 一般情形：共有 2n2n 人，其中 kk 位女性（题设 2<k<n2<k<n）。

总匹配数为 (2n−1)!!(2n-1)!!。

无女女对战：每位女性必须与男性配对。先从 2n−k2n-k 位男性中选出 kk 位与女性配对，并做一一匹配：共有

(2n−kk)⋅k!=(2n−k)!(2n−2k)!\binom{2n-k}{k}\cdot k!=\frac{(2n-k)!}{(2n-2k)!}

种；剩余 2n−2k2n-2k 位男性之间再任意配对，数目为 (2n−2k−1)!!(2n-2k-1)!!。

因此

p(k,n)=(2n−k)!(2n−2k)!⋅(2n−2k−1)!!(2n−1)!!.\boxed{p(k,n)=\frac{(2n-k)!}{(2n-2k)!}\cdot\frac{(2n-2k-1)!!}{(2n-1)!!}}.