Question

七支球队参加一项足球锦标赛，每支球队与其他每支球队恰好比赛一次。没有平局，每支球队在每场比赛中获胜的概率都是 $50\%$ ，且各场比赛结果相互独立。每场比赛中，胜者得 $1$ 分，负者得 $0$ 分。用总积分决定球队排名。在锦标赛第一场比赛中，球队 $A$ 击败球队 $B$ 。球队 $A$ 最终积分高于球队 $B$ 的概率为 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数。求 $m + n$ 。

Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded $1$ point and the loser gets $0$ points. The total points are accumulated to decide the ranks of the teams. In the first game of the tournament, team $A$ beats team $B.$ The probability that team $A$ finishes with more points than team $B$ is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：831

知识点：二项概率对称性组合

难度评级：2650

解答：

球队 $A$ 和 $B$ 各还剩 $5$ 场比赛，彼此之间没有比赛，所以所有 $2^5 \cdot 2^5 = 1024$ 种结果等可能。由于 $A$ 已领先一分， $A$ 最终积分更高恰好发生在 $A$ 剩余获胜场数至少与 $B$ 一样多时。

胜场数相等的结果数为由对称性，其余 $1024 - 252 = 772$ 种结果平均分为 $A$ 胜场更多和 $B$ 胜场更多两类。 $\sum_{k=0}^{5} \binom{5}{k}^2 = \binom{10}{5} = 252.$

所以概率为 $\frac{252 + 386}{1024} = \frac{638}{1024} = \frac{319}{512}$ ，且 $m + n = 319 + 512 = 831$ 。

Teams $A$ and $B$ each have $5$ games left, none against each other, so all $2^5 \cdot 2^5 = 1024$ outcomes are equally likely. Since $A$ already leads by one point, $A$ finishes with more points exactly when $A$ wins at least as many remaining games as $B$ does.

The number of outcomes with equal win counts is $\sum_{k=0}^{5} \binom{5}{k}^2 = \binom{10}{5} = 252.$ By symmetry, the other $1024 - 252 = 772$ outcomes split evenly between $A$ winning more and $B$ winning more.

So the probability is $\frac{252 + 386}{1024} = \frac{638}{1024} = \frac{319}{512},$ and $m + n = 319 + 512 = 831.$

2006 AIME II 第 10 题

答案：831

难度评级：2650

解答：

其他年份的第 10 题