2025 AMC 12B 第 18 题

Question

Awnik 反复玩一个获胜概率为

\dfrac{1}{3}

的游戏。各局结果相互独立。直到他至少赢过一次且输过一次为止，他所玩的局数的期望值是多少？

Awnik repeatedly plays a game that has a probability of winning of

\dfrac{1}{3}.

The outcomes of the games are independent. What is the expected value of the number of games he will play until he has both won and lost at least once?

\dfrac{5}{2}

3

\dfrac{16}{5}

\dfrac{7}{2}

\dfrac{15}{4}

答案：D

知识点：期望值几何分布

难度评级：1770

小提示：

第一局后已经出现了一种结果；之后只需要等待相反的结果。

After the first game you already have one outcome; you then wait only for the opposite outcome

大提示：

等待概率为

p

的结果首次出现的期望次数是

\dfrac{1}{p}

；按第一局是赢还是输分类。

The expected number of games to get an outcome of probability

p

is

\dfrac{1}{p};

condition on whether the first game is a win or a loss

解答：

第一局会产生一种结果。如果第一局赢了（概率

\tfrac{1}{3}

），则等待一次失败的期望局数为

\tfrac{1}{\frac{2}{3}} = \tfrac{3}{2}

；如果第一局输了（概率

\tfrac{2}{3}

），则等待一次获胜的期望局数为

\tfrac{1}{\frac{1}{3}} = 3

。所以总期望为

1 + \tfrac{1}{3}\cdot\tfrac{3}{2} + \tfrac{2}{3}\cdot 3 = 1 + \tfrac{1}{2} + 2

= \tfrac{7}{2}

。

所以正确答案是 D。

The first game produces one outcome. If it was a win (probability

\tfrac{1}{3}

), the expected wait for a loss is

\tfrac{1}{\frac{2}{3}} = \tfrac{3}{2};

if it was a loss (probability

\tfrac{2}{3}

), the expected wait for a win is

\tfrac{1}{\frac{1}{3}} = 3.

So the expected total is

1 + \tfrac{1}{3}\cdot\tfrac{3}{2} + \tfrac{2}{3}\cdot 3 = 1 + \tfrac{1}{2} + 2

= \tfrac{7}{2}.

Thus, the correct answer is D.

2025 AMC 12B 第 18 题

答案：D

难度评级：1770

小提示：

大提示：

解答：

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