2018 AMC 12A 第 24 题

Question

Alice、Bob 和 Carol 玩一个游戏，每人选择一个介于

0

和

1

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

0

和

1

之间的所有数中均匀随机选择一个数，Bob 宣布他会在

\tfrac12

和

\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

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

\tfrac12

and

\tfrac23.

Armed with this information, what number should Carol choose to maximize her chance of winning?

\tfrac12

\tfrac{13}{24}

\tfrac{7}{12}

\tfrac58

\tfrac23

答案：B

知识点：几何概率分类讨论最优化

难度评级：2520

小提示：

按 Carol 的数

c

相对于

\tfrac12

和

\tfrac23

的位置分类；当

\tfrac12 \lt c \lt \tfrac23

时她有两种获胜方式

Split into cases by where Carol’s number

c

sits relative to

\tfrac12

and

\tfrac23;

for

\tfrac12 \lt c \lt \tfrac23

she can win in two ways

大提示：

在中间区间内，她的获胜概率是

-12c^2 + 13c - 3

；在顶点

c = \tfrac{-b}{2a}

处最大化这个开口向下的二次式

In that middle range her win probability is

-12c^2 + 13c - 3;

maximize this downward quadratic at its vertex

c = \tfrac{-b}{2a}

解答：

若

c \le \tfrac12

，Carol 一定小于 Bob，所以她只有在 Alice 小于

c

时获胜，概率为

c \le \tfrac12

。若

c \ge \tfrac23

，她获胜的概率为

1 - c \le \tfrac13

。这两种情况都不超过

\tfrac12

。

对于

\tfrac12 \lt c \lt \tfrac23

，Bob 的数大于

c

的概率是

\frac{\frac{2}{3} - c}{\frac{2}{3} - \frac{1}{2}} = 4 - 6c

，所以 Carol 大于 Alice 且小于 Bob 的概率是

c(4 - 6c)

；反向排序的概率是

(1 - c)(6c - 3)

。相加得

\begin{aligned} &c(4 - 6c) + (1 - c)(6c - 3) \\ &= -12c^2 + 13c - 3 \end{aligned}\text{。}

这个开口向下的抛物线在

c = \tfrac{13}{24}

处取到最大值，该点位于

\left(\tfrac12, \tfrac23\right)

内，且最大值超过

\tfrac12

。

所以正确答案是 B。

If

c \le \tfrac12,

Carol beats Bob automatically, so she wins only if Alice is below

c,

probability

c \le \tfrac12.

If

c \ge \tfrac23,

she wins with probability

1 - c \le \tfrac13.

Neither case exceeds

\tfrac12.

For

\tfrac12 \lt c \lt \tfrac23,

the chance Bob’s number exceeds

c

is

\frac{\frac{2}{3} - c}{\frac{2}{3} - \frac{1}{2}} = 4 - 6c,

so the probability Carol is above Alice and below Bob is

c(4 - 6c);

the reverse ordering has probability

(1 - c)(6c - 3).

Adding,

\begin{aligned} &c(4 - 6c) + (1 - c)(6c - 3) \\ &= -12c^2 + 13c - 3. \end{aligned}

This downward parabola is maximized at

c = \tfrac{13}{24},

which lies in

\left(\tfrac12, \tfrac23\right),

and its value exceeds

\tfrac12.

Thus, the correct answer is B.

2018 AMC 12A 第 24 题

答案：B

难度评级：2520

小提示：

大提示：

解答：

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