AMC12 2025 B

AMC12 2025 B · Q18

AMC12 2025 B · Q18. It mainly tests Probability (basic), Expected value (basic).

Awnik repeatedly plays a game that has a probability of winning of $\frac{1}{3}$. 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?

Awnik反复玩一个获胜概率为$\frac{1}{3}$的游戏。各游戏结果独立。他玩到既赢过又输过至少一次的游戏数期望值为多少？

(A) \frac{5}{2} \frac{5}{2}

(B) 3 3

(D) \frac{7}{2} \frac{7}{2}

(E) \frac{15}{4} \frac{15}{4}

Answer

Correct choice: (D)

正确答案：（D）

Solution

Let the probability of a win be $p$ and the probability of a loss be $q$. If the first game is a win, then we must find the expected number of further games to get a loss, which will be $\frac{1}{q}$. In addition to the first game played, there will be $1+\frac{1}{q}$ games played. Therefore, the expected number of games needed to get a win on game $1$ and then a loss is $(p)\left(1+\frac{1}{q}\right).$ Similarly, if the first game is a loss, we need to the find the expected number of further games to get a win, which will be $\frac{1}{p}$. There will be a total of $1+\frac{1}{p}$ games played. Therefore, the expected number of games needed to get a loss on game $1$ and then a win is $(q)\left(1+\frac{1}{p}\right).$ The answer is \[(p)\left(1+\frac{1}{q}\right) + (q)\left(1+\frac{1}{p}\right)\] \[\left(\frac{1}{3}\right) \left(1+\frac{3}{2}\right) + \left(\frac{2}{3}\right)(1+3) = \frac{5}{6} + \frac{8}{3} = \boxed{\frac{7}{2}}.\]

设获胜概率为$p$，失败概率为$q$。若第一局赢，则需进一步求得失败的期望游戏数为$\frac{1}{q}$。加上第一局，共$1+\frac{1}{q}$局。故第一局赢后总期望为$(p)\left(1+\frac{1}{q}\right)$。类似，若第一局输，则需进一步求得获胜的期望游戏数为$\frac{1}{p}$，总共$1+\frac{1}{p}$局。故第一局输后总期望为$(q)\left(1+\frac{1}{p}\right)$。答案为 \[(p)\left(1+\frac{1}{q}\right) + (q)\left(1+\frac{1}{p}\right)\] \[\left(\frac{1}{3}\right) \left(1+\frac{3}{2}\right) + \left(\frac{2}{3}\right)(1+3) = \frac{5}{6} + \frac{8}{3} = \boxed{\frac{7}{2}}.\]

Topics