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四个密封箱：公平开箱费 X

four sealed boxes

专题
Strategy / 策略
难度
L4

题目详情

量化面试题：四个密封箱：公平开箱费 X。

英文原题

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 XX，其中 1 个箱子有 100，其余为 0。

用动态规划做最优停：令 VnV_n 为当还剩 nn 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

Vn=max⁡{0, −X+100n+n−1nVn−1},V0=0.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=4n=4 时应满足 V4=0V_4=0。

猜测最优策略会一直开到找到 100（或开完），因此 V1=100−XV_1=100-X，递推得到

V2=−X+50+12(100−X)=100−32X,V_2=-X+50+\frac12(100-X)=100-\frac{3}{2}X, V3=−X+1003+23V2=100−116X,V_3=-X+\frac{100}{3}+\frac{2}{3}V_2=100-\frac{11}{6}X, V4=−X+25+34V3=100−52X.V_4=-X+25+\frac34 V_3=100-\frac{5}{2}X.

令 V4=0V_4=0 得

X=40.\boxed{X=40}.

英文解析

Vn=max⁡{0, −X+100n+n−1nVn−1},V0=0.V_n=\max\left\{0,\ -X+\frac{100}{n}+\frac{n-1}{n}V_{n-1}\right\},\quad V_0=0. V2=−X+50+12(100−X)=100−32X,V_2=-X+50+\frac12(100-X)=100-\frac{3}{2}X, V3=−X+1003+23V2=100−116X,V_3=-X+\frac{100}{3}+\frac{2}{3}V_2=100-\frac{11}{6}X, V4=−X+25+34V3=100−52X.V_4=-X+25+\frac34 V_3=100-\frac{5}{2}X. X=40.\boxed{X=40}.