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2017 AMC 12B 第 22 题

先试着解答 2017 AMC 12B 第 22 题，然后核对你的答案与精心整理的解答，解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2017 AMC 12B 解答，或核对答案。

所有题目均经美国数学协会（MAA）官方合法授权使用。

22.

Abby、Bernardo、Carl 和 Debra 玩一个游戏，每个人开始时都有四枚硬币。游戏共四轮。每一轮，把四个球放入一个瓮中：一个绿色、一个红色、两个白色。玩家们依次随机抽球且不放回。抽到绿球的人给抽到红球的人一枚硬币。第四轮结束时，每位玩家都有四枚硬币的概率是多少？

Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn—one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins?

7576\dfrac{7}{576}

5192\dfrac{5}{192}

136\dfrac{1}{36}

5144\dfrac{5}{144}

748\dfrac{7}{48}

答案：B
知识点：基本概率分类讨论
难度评级：2330
小提示：

每轮有 4⋅3=124 \cdot 3 = 12 个等可能的（给出者，接收者）结果，所以共有 12412^4 个结果序列

Each round has 4⋅3=124 \cdot 3 = 12 equally likely (giver, receiver) outcomes, for 12412^4 total

大提示：

统计能让每个人回到 44 枚硬币的结果序列：每个人的净转移必须抵消；按交换模式分类

Count outcome sequences that return everyone to 44 coins: each player’s net transfers must cancel; split into cases by the pattern of exchanges

解答：

每轮有 4⋅3=124 \cdot 3 = 12 个等可能的（给出者，接收者）有序对，所以共有 12412^4 个结果序列。每个人最后都有四枚硬币，当且仅当四次转移相互抵消。有利模式为：一次 44-循环赠送（24⋅6=14424 \cdot 6 = 144 种），两组不相交的相互交换（24⋅3=7224 \cdot 3 = 72 种），同一对玩家互相交换两次（6⋅6=366 \cdot 6 = 36 种），以及某个玩家既给另外两人各一枚又从这两人各收一枚（4⋅3⋅24=2884 \cdot 3 \cdot 24 = 288 种）。总数为 144+72+36+288=540144 + 72 + 36 + 288 = 540，概率为 540124=54020736=5192\dfrac{540}{12^4} = \dfrac{540}{20736} = \dfrac{5}{192}。

所以正确答案是 B。

Each round has 4⋅3=124 \cdot 3 = 12 equally likely (giver, receiver) pairs, so there are 12412^4 outcome sequences. Everyone ends with four coins exactly when the four transfers cancel. The favorable patterns are: a 44-cycle of gifts (24⋅6=14424 \cdot 6 = 144 ways), two disjoint mutual exchanges (24⋅3=7224 \cdot 3 = 72), one pair exchanging twice (6⋅6=366 \cdot 6 = 36), and one player both giving to and receiving from each of two others (4⋅3⋅24=2884 \cdot 3 \cdot 24 = 288). These total 144+72+36+288=540.144 + 72 + 36 + 288 = 540. The probability is 540124=54020736=5192.\dfrac{540}{12^4} = \dfrac{540}{20736} = \dfrac{5}{192}.

Thus, the correct answer is B.

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