2019 AMC 12B 第 19 题
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19.
Raashan、Sylvia 和 Ted 玩下面的游戏。每人初始有 。 每隔 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 。铃响 次后,每个玩家都有 的概率是多少? (例如,Raashan 和 Ted 可以都决定给 Sylvia 而 Sylvia 可以决定把她的一美元给 Ted, 这时 Raashan 有 , Sylvia 有 , Ted 有 , 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 则第二轮结束时持有金额不变。)
Raashan, Sylvia, and Ted play the following game. Each starts with A bell rings every 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 to that player. What is the probability that after the bell has rung times, each player will have (For example, Raashan and Ted may each decide to give to Sylvia, and Sylvia may decide to give her dollar to Ted, at which point Raashan will have Sylvia will have and Ted will have 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 to, and the holdings will be the same at the end of the second round.)
答案:B
解答:
从 , 出发,三名玩家各自把钱给另外两人之一,所以有 个等可能结果; 只有 种循环赠送模式会回到 ,概率为 。
从 状态出发,没钱的玩家不给钱;检查另外两人的 个等可能选择,恰好一个会得到 ,概率仍为 。
因此任意一次铃响后,状态为 的概率都是 , 包括铃响 次之后。
所以正确答案是 B。
From each of the three players gives to one of two others, so there are equally likely outcomes; only the cyclic gift patterns return to a probability of
From a state the broke player gives nothing, and checking the equally likely choices of the other two shows exactly one yields again probability
So after any ring the probability of is including after rings.
Thus, B is the correct answer.
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