AMC12 2019 B

AMC12 2019 B · Q19

AMC12 2019 B · Q19. It mainly tests Probability (basic), Games (basic).

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元。每15秒铃声响起，此时每个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元。铃声响起2019次后，每人仍有1元的概率是多少？（例如，Raashan和Ted可能都决定给Sylvia 1元，而Sylvia决定给Ted她的1元，此时Raashan有0元，Sylvia有2元，Ted有1元，第一轮结束。第二轮Raashan没钱给，但Sylvia和Ted可能互相给1元，持有可能与结束第二轮时相同。）

(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某一张美元给出去。因此，金钱分配只有两种可能：1-1-1 和 2-1-0。假设游戏某一轮从 1-1-1 开始。不失一般性，设 Raashan 把他的 1 美元给 Sylvia。此时这一轮要以 1-1-1 结束，唯一的方法是：Ted 把他的 1 美元给 Raashan（否则 Sylvia 会变成 \$2），并且 Sylvia 把她的 1 美元给 Ted；其概率为 $\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$。再假设某一轮从 2-1-0 开始；不失一般性，设 Raashan 有 \$2，Sylvia 有 \$1。此时这一轮要以 1-1-1 结束，唯一的方法是：Sylvia 把她的 1 美元给 Ted（否则 Raashan 会变成 \$2），并且 Raashan 把他的 1 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，这一轮以 1-1-1 结束的概率都是 $\frac{1}{4}$。特别地，在第 2019 轮结束时，每位玩家都拥有 \$1 的概率为 $\frac{1}{4}$。

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