Question

九张牌分别编号为 $1, 2, 3, \ldots, 9$ ，三名玩家各自随机选择并保留三张牌，然后求自己三张牌上的数之和。三名玩家所得和全为奇数的概率为 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数。求 $m + n$ 。

Nine tiles are numbered $1, 2, 3, \ldots, 9,$ respectively. Each of three players randomly selects and keeps three of the tiles, and sums those three values. The probability that all three players obtain an odd sum is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：17

知识点：基本概率奇偶性组合

难度评级：2350

解答：

一名玩家的三张牌之和为奇数，当且仅当他拿到奇数张奇数牌，也就是一张或三张。九张牌中有五张奇数牌和四张偶数牌，把五张奇数牌分到三个组中且每组奇数牌数为一或三，唯一的分法类型是 $3 + 1 + 1$ 。

计算有利的发牌方式：选择哪名玩家得到三张奇数牌，有 $3$ 种；选择这名玩家的三张奇数牌，有 $\binom{5}{3} = 10$ 种；把剩下两张奇数牌各给另外两名玩家，有 $2$ 种；再把四张偶数牌按二二分给这两名玩家，有 $\binom{4}{2} = 6$ 种。因此有利发牌方式为 $3 \cdot 10 \cdot 2 \cdot 6 = 360$ 种。总发牌方式为 $\binom{9}{3}\binom{6}{3} = 84 \cdot 20 = 1680$ 种。

概率为 $\frac{360}{1680} = \frac{3}{14}$ ，所以 $m + n = 3 + 14 = 17$ 。

A player's three tiles have an odd sum exactly when the player holds an odd number of odd tiles — one or three. The nine tiles include five odd and four even, and the only way to split five odd tiles into three groups of size one or three is $3 + 1 + 1.$

Count favorable deals: choose which player gets three odd tiles ( $3$ ways), choose that player's odd tiles ( $\binom{5}{3} = 10$ ways), give one of the two remaining odd tiles to each other player ( $2$ ways), then split the four even tiles two and two between those players ( $\binom{4}{2} = 6$ ways), for $3 \cdot 10 \cdot 2 \cdot 6 = 360$ deals. The total number of deals is $\binom{9}{3}\binom{6}{3} = 84 \cdot 20 = 1680.$

The probability is $\frac{360}{1680} = \frac{3}{14},$ so $m + n = 3 + 14 = 17.$

1998 AIME 第 4 题

答案：17

难度评级：2350

解答：

其他年份的第 4 题