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：

其他年份的第 18 题