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2018 AMC 12A 第 24 题

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所有题目均经美国数学协会（MAA）官方合法授权使用。

24.

Alice、Bob 和 Carol 玩一个游戏，每人选择一个介于 00 和 11 之间的实数。游戏获胜者是其数字位于另外两名玩家所选数字之间的人。Alice 宣布她会在 00 和 11 之间的所有数中均匀随机选择一个数，Bob 宣布他会在 12\tfrac12 和 23\tfrac23 之间的所有数中均匀随机选择一个数。在知道这些信息后，Carol 应该选择什么数来最大化她获胜的概率？

Alice, Bob, and Carol play a game in which each of them chooses a real number between 00 and 1.1. 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 00 and 1,1, and Bob announces that he will choose his number uniformly at random from all the numbers between 12\tfrac12 and 23.\tfrac23. Armed with this information, what number should Carol choose to maximize her chance of winning?

12\tfrac12

1324\tfrac{13}{24}

712\tfrac{7}{12}

58\tfrac58

23\tfrac23

答案：B
知识点：几何概率分类讨论最优化
难度评级：2520
解答：

若 c≤12c \le \tfrac12，Carol 一定小于 Bob，所以她只有在 Alice 小于 cc 时获胜，概率为 c≤12c \le \tfrac12。若 c≥23c \ge \tfrac23，她获胜的概率为 1−c≤131 - c \le \tfrac13。这两种情况都不超过 12\tfrac12。

对于 12<c<23\tfrac12 \lt c \lt \tfrac23，Bob 的数大于 cc 的概率是 2/3−c2/3−1/2=4−6c\frac{2/3 - c}{2/3 - 1/2} = 4 - 6c，所以 Carol 大于 Alice 且小于 Bob 的概率是 c(4−6c)c(4 - 6c)；反向排序的概率是 (1−c)(6c−3)(1 - c)(6c - 3)。相加得 这个开口向下的抛物线在 c=1324c = \tfrac{13}{24} 处取最大值，该值位于 (12,23)\left(\tfrac12, \tfrac23\right) 内，且最大值超过 12\tfrac12。 c(4−6c)+(1−c)(6c−3)=−12c2+13c−3. \begin{aligned} &c(4 - 6c) + (1 - c)(6c - 3) \\ &= -12c^2 + 13c - 3. \end{aligned}

所以正确答案是 B。

If c≤12,c \le \tfrac12, Carol beats Bob automatically, so she wins only if Alice is below c,c, probability c≤12.c \le \tfrac12. If c≥23,c \ge \tfrac23, she wins with probability 1−c≤13.1 - c \le \tfrac13. Neither case exceeds 12.\tfrac12.

For 12<c<23,\tfrac12 \lt c \lt \tfrac23, the chance Bob's number exceeds cc is 2/3−c2/3−1/2=4−6c,\frac{2/3 - c}{2/3 - 1/2} = 4 - 6c, so the probability Carol is above Alice and below Bob is c(4−6c);c(4 - 6c); the reverse ordering has probability (1−c)(6c−3).(1 - c)(6c - 3). Adding, c(4−6c)+(1−c)(6c−3)=−12c2+13c−3. \begin{aligned} &c(4 - 6c) + (1 - c)(6c - 3) \\ &= -12c^2 + 13c - 3. \end{aligned} This downward parabola is maximized at c=1324,c = \tfrac{13}{24}, which lies in (12,23),\left(\tfrac12, \tfrac23\right), and its value exceeds 12.\tfrac12.

Thus, the correct answer is B.

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