AMC10 2019 B

AMC10 2019 B · Q22

AMC10 2019 B · Q22. It mainly tests Probability (basic), Invariants.

Raashan, Sylvia, and Ted play the following game. Each starts with $1$. A bell rings every 15 seconds, at which time each of the players who currently has money simultaneously chooses one of the other two players independently and at random and gives $1$ to that player. What is the probability that after the bell has rung 2019 times, each player will have $1$? (For example, Raashan and Ted may each decide to give $1$ to Sylvia, and Sylvia may decide to give her dollar to Ted, at which point Raashan will have $0$, Sylvia will have $2$, and Ted will have $1$, 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 $1$ to, and the holdings will be the same at the end of the second round.)

Raashan、Sylvia 和 Ted 玩以下游戏。每人起始有 $1$。每 15 秒铃声响起一次，此时每个目前有钱的玩家同时独立随机选择其他两个玩家中的一个，并给该玩家 $1$。铃声响起 2019 次后，每位玩家都有 $1$ 的概率是多少？（例如，Raashan 和 Ted 可能都决定给 $1$ 给 Sylvia，而 Sylvia 可能决定把她的 $1$ 给 Ted，此时 Raashan 有 $0$，Sylvia 有 $2$，Ted 有 $1$，第一轮结束。第二轮 Raashan 没有钱给，但 Sylvia 和 Ted 可能选择互相给 $1$，第二轮结束时持有量相同。）

(A) \frac{1}{7} \frac{1}{7}

(B) \frac{1}{4} \frac{1}{4}

(D) \frac{1}{2} \frac{1}{2}

(E) \frac{2}{3} \frac{2}{3}

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): No player can ever end up with \$3 at the end of a round, because that player had to give away one of the dollars in play. Therefore the only two possible distributions of the money are 1-1-1 and 2-1-0. Suppose that a round of the game starts at 1-1-1. Without loss of generality, assume that Raashan gives his dollar to Sylvia. Then the only way for the round to end at 1-1-1 is for Ted to give his dollar to Raashan (otherwise Sylvia would end up with \$2) and for Sylvia to give her dollar to Ted; the probability of this is $\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$. Next suppose that a round starts at 2-1-0; without loss of generality, assume that Raashan has \$2 and Sylvia has \$1. Then the only way for the round to end at 1-1-1 is for Sylvia to give her dollar to Ted (otherwise Raashan would end up with \$2) and for Raashan to give his dollar to Sylvia; the probability of this is $\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$. Thus no matter how the round starts, the probability that the round will end at 1-1-1 is $\frac{1}{4}$. In particular, the probability is $\frac{1}{4}$ that at the end of the 2019th round each player will have \$1.

答案（B）：在一轮结束时，没有任何玩家可能最终拥有 \$3，因为该玩家必须把场上流通的其中一美元送出。因此，金钱分配只有两种可能：1-1-1 和 2-1-0。假设游戏的一轮从 1-1-1 开始。不失一般性，设 Raashan 把他的一美元给 Sylvia。则这一轮以 1-1-1 结束的唯一方式是：Ted 把他的一美元给 Raashan（否则 Sylvia 将会有 \$2），并且 Sylvia 把她的一美元给 Ted；其概率为 $\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$。再假设一轮从 2-1-0 开始；不失一般性，设 Raashan 有 \$2 而 Sylvia 有 \$1。则这一轮以 1-1-1 结束的唯一方式是：Sylvia 把她的一美元给 Ted（否则 Raashan 将会有 \$2），并且 Raashan 把他的一美元给 Sylvia；其概率为 $\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$。因此，无论这一轮如何开始，它以 1-1-1 结束的概率都是 $\frac{1}{4}$。特别地，第 2019 轮结束时每位玩家都有 \$1 的概率为 $\frac{1}{4}$。

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