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2019 AMC 12B 第 19 题

先试着解答 2019 AMC 12B 第 19 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 2019 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。

19.

Raashan、Sylvia 和 Ted 玩下面的游戏。每人初始有 $1\$1。每隔 1515 秒铃响一次，此时当前有钱的每个玩家 同时独立随机地选择另外两名玩家之一，并给该玩家 $1\$1。铃响 20192019 次后，每个玩家都有 $1\$1 的概率是多少？（例如，Raashan 和 Ted 可以都决定给 Sylvia $1\$1，而 Sylvia 可以决定把她的一美元给 Ted，这时 Raashan 有 $0\$0，Sylvia 有 $2\$2，Ted 有 $1\$1，第一轮游戏结束。第二轮中 Raashan 没有钱可给，但 Sylvia 和 Ted 可能互相选择给对方 $1\$1，则第二轮结束时持有金额不变。）

Raashan, Sylvia, and Ted play the following game. Each starts with $1.\$1. A bell rings every 1515 seconds, at which time each of the players who currently have money simultaneously chooses one of the other two players independently and at random and gives $1\$1 to that player. What is the probability that after the bell has rung 20192019 times, each player will have $1?\$1? (For example, Raashan and Ted may each decide to give $1\$1 to Sylvia, and Sylvia may decide to give her dollar to Ted, at which point Raashan will have $0,\$0, Sylvia will have $2,\$2, and Ted will have $1,\$1, and that is the end of the first round of play. In the second round Raashan has no money to give, but Sylvia and Ted might choose each other to give their $1\$1 to, and the holdings will be the same at the end of the second round.)

17\dfrac{1}{7}

14\dfrac{1}{4}

13\dfrac{1}{3}

12\dfrac{1}{2}

23\dfrac{2}{3}

答案：B
知识点：递推概率对称性不变量
难度评级：1980
小提示：

可能的状态只有 (1,1,1)(1,1,1) 和 (2,1,0)(2,1,0) 的排列

The only states are (1,1,1)(1,1,1) and permutations of (2,1,0)(2,1,0)

大提示：

证明从任一种状态出发，下一状态为 (1,1,1)(1,1,1) 的概率都是 14\dfrac14。

Show that from either kind of state the next state is (1,1,1)(1,1,1) with probability 14\dfrac14

解答：

从 (1,1,1)(1,1,1) 出发，三名玩家各自把钱给另外两人之一，所以有 88 个等可能结果；只有 22 种循环赠送模式会回到 (1,1,1)(1,1,1)，概率为 14\dfrac14。

从 (2,1,0)(2,1,0) 状态出发，没钱的玩家不给钱；检查另外两人的 44 个等可能选择，恰好一个会得到 (1,1,1)(1,1,1)，概率仍为 14\dfrac14。

因此任意一次铃响后，状态为 (1,1,1)(1,1,1) 的概率都是 14\dfrac14，包括铃响 20192019 次之后。

所以正确答案是 B。

From (1,1,1),(1,1,1), each of the three players gives to one of two others, so there are 88 equally likely outcomes; only the 22 cyclic gift patterns return to (1,1,1),(1,1,1), a probability of 14.\dfrac14.

From a (2,1,0)(2,1,0) state the broke player gives nothing, and checking the 44 equally likely choices of the other two shows exactly one yields (1,1,1),(1,1,1), again probability 14.\dfrac14.

So after any ring the probability of (1,1,1)(1,1,1) is 14,\dfrac14, including after 20192019 rings.

Thus, B is the correct answer.

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