Question

Raashan、Sylvia 和 Ted 玩下面的游戏。每人起初有 $\$1$ 。每 $15$ 秒铃响一次，此时每个当前有钱的玩家都同时、独立且随机地选择另外两名玩家之一，并给那人 $\$1$ 。铃响 $2019$ 次后，每个玩家都有 $\$1$ 的概率是多少？

例如，Raashan 和 Ted 可以都决定给 Sylvia 一美元，而 Sylvia 决定把她的 $\$1$ 给 Ted；这时 Raashan 有 $\$0$ ，Sylvia 有 $\$2$ ，Ted 有 $\$1$ ，第一轮结束。第二轮 Raashan 没钱可给，但 Sylvia 和 Ted 可能互相给对方 $\$1$ ，结果钱数保持不变。

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.)

$\dfrac{1}{7}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

答案：B

知识点：递推概率对称性

难度评级：1950

解答：

不计顺序，唯一能达到的钱数状态是 $(1,1,1)$ 和 $(2,1,0)$ 。玩家不可能在一轮结束时拥有全部 $3$ 美元：每个轮初有钱的人都必须给别人一美元，而且不能给自己。从 $(1,1,1)$ 出发，下一状态仍为 $(1,1,1)$ ，当且仅当三个人沿同一个循环方向传钱，其概率为 $2\left(\dfrac12\right)^3=\dfrac14$ 。

从 $(2,1,0)$ 出发，记三人的钱数为 $A=2,B=1,C=0$ 。下一状态恰为 $(1,1,1)$ ，当且仅当 $A$ 把钱给 $B$ ，且 $B$ 把钱给 $C$ ，这是四个等可能选择组合中的一个。因此概率同样是 $\dfrac14$ 。

所以无论第 $2018$ 次铃响后的状态如何，下一次铃响后状态为 $(1,1,1)$ 的概率都是 $\dfrac14$ 。所以正确答案是 B。

The only reachable money configurations up to order are $(1,1,1)$ and $(2,1,0)$ . A player cannot finish a round with all $3$ dollars: anyone who begins with money must give a dollar to someone else, and no one can give to themselves. From $(1,1,1)$ , the next state is again $(1,1,1)$ exactly when all three players pass dollars in the same cyclic direction, which has probability $2\left(\dfrac12\right)^3=\dfrac14$ .

From $(2,1,0)$ , label the players' holdings $A=2,B=1,C=0$ . The next state is $(1,1,1)$ exactly when $A$ gives to $B$ and $B$ gives to $C$ , one of the four equally likely pairs of choices. This also has probability $\dfrac14$ .

Therefore, regardless of the state after $2018$ rings, the probability that the state after the next ring is $(1,1,1)$ is $\dfrac14$ . Thus, B is the correct answer.

2019 AMC 10B 第 22 题

答案：B

难度评级：1950

解答：

其他年份的第 22 题