AMC12 2018 A

AMC12 2018 A · Q24

AMC12 2018 A · Q24. It mainly tests Probability (basic), Counting & probability misc.

Alice, Bob, and Carol play a game in which each of them chooses a real number between 0 and 1. The winner of the game is the one whose number is between the numbers chosen by the other two players. Alice announces that she will choose her number uniformly at random from all the numbers between 0 and 1, and Bob announces that he will choose his number uniformly at random from all the numbers between $\frac{1}{2}$ and $\frac{2}{3}$. Armed with this information, what number should Carol choose to maximize her chance of winning?

Alice、Bob 和 Carol 玩一个游戏，每人选择 0 到 1 之间的实数。获胜者是其数位于其他两人所选数之间的那个人。Alice 宣布她将从 0 到 1 的所有数中均匀随机选择她的数，Bob 宣布他将从 $\frac{1}{2}$ 到 $\frac{2}{3}$ 的所有数中均匀随机选择他的数。有了这些信息，Carol 应该选择什么数来最大化她的获胜概率？

(A) $\frac{1}{2}$ $\frac{1}{2}$

(B) $\frac{13}{24}$ $\frac{13}{24}$

(D) $\frac{5}{8}$ $\frac{5}{8}$

(E) $\frac{2}{3}$ $\frac{2}{3}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Because Alice and Bob are choosing their numbers uniformly at random, the cases in which two or three of the chosen numbers are equal have probability $0$ and can be ignored. Suppose Carol chooses the number $c$. She will win if her number is greater than Alice’s number and less than Bob’s, and she will win if her number is less than Alice’s number and greater than Bob’s. There are three cases. • If $c \le \frac{1}{2}$, then Carol’s number is automatically less than Bob’s, so her chance of winning is the probability that Alice’s number is less than $c$, which is just $c$. The best that Carol can do in this case is to choose $c=\frac{1}{2}$, in which case her chance of winning is $\frac{1}{2}$. • If $c \ge \frac{2}{3}$, then Carol’s number is automatically greater than Bob’s, so her chance of winning is the probability that Alice’s number is greater than $c$, which is just $1-c$. The best that Carol can do in this case is to choose $c=\frac{2}{3}$, in which case her chance of winning is $\frac{1}{3}$. • Finally suppose that $\frac{1}{2}<c<\frac{2}{3}$. The probability that Carol’s number is less than Bob’s is $$ \frac{\frac{2}{3}-c}{\frac{2}{3}-\frac{1}{2}}=4-6c, $$ so the probability that her number is greater than Alice’s and less than Bob’s is $c(4-6c)$. Similarly, the probability that her number is less than Alice’s and greater than Bob’s is $(1-c)(6c-3)$. Carol’s probability of winning in this case is therefore $$ c(4-6c)+(1-c)(6c-3)=-12c^2+13c-3. $$ The value of a quadratic polynomial with a negative coefficient on its quadratic term is maximized at $-\frac{b}{2a}$, where $a$ is the coefficient on its quadratic term and $b$ is the coefficient on its linear term; here that is when $c=\frac{13}{24}$, which is indeed between $\frac{1}{2}$ and $\frac{2}{3}$. Her probability of winning is then $$ -12\cdot\left(\frac{13}{24}\right)^2+13\cdot\frac{13}{24}-3=\frac{25}{48}>\frac{24}{48}=\frac{1}{2}. $$ Because the probability of winning in the third case exceeds the probabilities obtained in the first two cases, Carol should choose $\frac{13}{24}$.

答案（B）：因为 Alice 和 Bob 以均匀随机方式选数，所以出现两个或三个所选数字相等的情况其概率为 $0$，可以忽略。设 Carol 选择数字 $c$。当她的数字大于 Alice 的数字且小于 Bob 的数字时她会赢；当她的数字小于 Alice 的数字且大于 Bob 的数字时她也会赢。分三种情况。 • 若 $c \le \frac{1}{2}$，则 Carol 的数必然小于 Bob 的数，因此她获胜的概率等于 Alice 的数小于 $c$ 的概率，也就是 $c$。此时 Carol 能做的最好选择是 $c=\frac{1}{2}$，此时她的获胜概率为 $\frac{1}{2}$。 • 若 $c \ge \frac{2}{3}$，则 Carol 的数必然大于 Bob 的数，因此她获胜的概率等于 Alice 的数大于 $c$ 的概率，也就是 $1-c$。此时 Carol 能做的最好选择是 $c=\frac{2}{3}$，此时她的获胜概率为 $\frac{1}{3}$。 • 最后设 $\frac{1}{2}<c<\frac{2}{3}$。Carol 的数小于 Bob 的数的概率为 $$ \frac{\frac{2}{3}-c}{\frac{2}{3}-\frac{1}{2}}=4-6c, $$ 因此她的数大于 Alice 且小于 Bob 的概率为 $c(4-6c)$。同理，她的数小于 Alice 且大于 Bob 的概率为 $(1-c)(6c-3)$。所以在这种情况下 Carol 的获胜概率为 $$ c(4-6c)+(1-c)(6c-3)=-12c^2+13c-3. $$ 二次项系数为负的二次多项式在 $-\frac{b}{2a}$ 处取得最大值，其中 $a$ 为二次项系数、$b$ 为一次项系数；这里对应 $c=\frac{13}{24}$，确实介于 $\frac{1}{2}$ 与 $\frac{2}{3}$ 之间。此时她的获胜概率为 $$ -12\cdot\left(\frac{13}{24}\right)^2+13\cdot\frac{13}{24}-3=\frac{25}{48}>\frac{24}{48}=\frac{1}{2}. $$ 由于第三种情况的获胜概率超过前两种情况得到的概率，Carol 应选择 $\frac{13}{24}$。

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