Question

Truncator 俱乐部与另外六支队同在一个足球联赛中，并与每支队比赛一次。在它的 $6$ 场比赛中， Truncator 俱乐部获胜、失利、打平的概率各为 $\frac{1}{3}$ 。Truncator 俱乐部赛季结束时胜场数多于负场数的概率为 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数。求 $m + n$ 。

Club Truncator is in a soccer league with six other teams, each of which it plays once. In any of its $6$ matches, the probabilities that Club Truncator will win, lose, or tie are each $\frac{1}{3}.$ The probability that Club Truncator will finish the season with more wins than losses is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：341

知识点：基本概率多重集排列补集计数对称性

难度评级：2560

解答：

交换胜负是保持概率不变的对称操作，所以胜场多于负场的概率 $P$ 等于负场多于胜场的概率。因此 $P = \frac{1 - p_0}{2}$ ，其中 $p_0$ 是胜场数和负场数相等的概率。

若有 $k$ 场胜、 $k$ 场负、 $6 - 2k$ 场平，则这些结果可排列成 $\frac{6!}{k! \, k! \, (6 - 2k)!}$ 种。当 $k = 0, 1, 2, 3$ 时，这些数量分别为 $1$ 、 $30$ 、 $90$ 、 $20$ ，总计 $141$ 。所有 $3^6 = 729$ 个结果序列等可能，所以 $p_0 = \frac{141}{729} = \frac{47}{243}$ 。

因此 $P = \frac{1}{2}\left(1 - \frac{47}{243}\right) = \frac{98}{243}$ ，所以 $m + n = 98 + 243 = 341$ 。

Swapping wins and losses is a probability-preserving symmetry, so the probability $P$ of more wins than losses equals the probability of more losses than wins, giving $P = \frac{1 - p_0}{2},$ where $p_0$ is the probability of equally many wins and losses.

An outcome with $k$ wins, $k$ losses, and $6 - 2k$ ties can be arranged in $\frac{6!}{k! \, k! \, (6 - 2k)!}$ ways: $1,$ $30,$ $90,$ $20$ for $k = 0, 1, 2, 3,$ totaling $141.$ Each of the $3^6 = 729$ outcome sequences is equally likely, so $p_0 = \frac{141}{729} = \frac{47}{243}.$

Therefore $P = \frac{1}{2}\left(1 - \frac{47}{243}\right) = \frac{98}{243},$ and $m + n = 98 + 243 = 341.$

2001 AIME II 第 11 题

答案：341

难度评级：2560

解答：

其他年份的第 11 题