四个密封箱：公平开箱费 X

four sealed boxes

专题: Strategy / 策略
难度: L4
来源: QuantQuestion

题目详情

Let's play a game. There are four sealed boxes. There is 100 pounds in one box and the others are empty. A player can open a box and take the contents as many times as they like, but each time they do so they must pay X. Assuming this is a fair game, what is the value of X?

解析

设每次开一个尚未打开的箱子要付 $X$ ，其中 1 个箱子有 100，其余为 0。

用动态规划做最优停：令 $V_n$ 为当还剩 $n$ 个未开箱时的最优期望净收益（允许随时停止，收益 0）。

若选择再开 1 个箱子，则

V_n=\max\left\{0,\ -X+\frac{100}{n}+\frac{n-1}{n}V_{n-1}\right\},\quad V_0=0.

在公平游戏下，起始 $n=4$ 时应满足 $V_4=0$ 。

猜测最优策略会一直开到找到 100（或开完），因此 $V_1=100-X$ ，递推得到

V_2=-X+50+\frac12(100-X)=100-\frac{3}{2}X,

V_3=-X+\frac{100}{3}+\frac{2}{3}V_2=100-\frac{11}{6}X,

V_4=-X+25+\frac34 V_3=100-\frac{5}{2}X.

令 $V_4=0$ 得

\boxed{X=40}.