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30 面骰对 20 面骰

30 Die Split III

专题
Probability / 概率
难度
L6

题目详情

Alice 有一枚公平 30 面骰，Bob 有一枚公平 20 面骰。两人同时掷骰，点数更大者胜；若平局则 Bob 胜。

Bob 掷完后可以选择是否重掷一次（他看不到 Alice 的点数），若选择重掷则用第二次点数作为最终点数。

在 Bob 最优策略下，求 Alice 获胜概率。

Alice and Bob have fair 3030-sided and 2020-sided dice, respectively. The goal for each player is to have the largest value on their die. Alice and Bob both roll their dice. However, Bob has the option to re-roll his die in the event that he is unhappy with the outcome. He can't see Alice's die beforehand. Bob then keeps the value of the new die roll. In the event of a tie, Bob is the winner. Assuming optimal play by Bob, find the probability Alice is the winner.

解析

设 Alice 点数为 AA，Bob 最终点数为 BB。因为 A∼Unif{1,…,30}A\sim\mathrm{Unif}\{1,\dots,30\}，有

P(A>B)=E[30−B30]=1−E[B]30.P(A>B)=\mathbb{E}\left[\frac{30-B}{30}\right]=1-\frac{\mathbb{E}[B]}{30}.

因此 Bob 应最大化 E[B]\mathbb{E}[B]。

若 Bob 采用阈值策略：第一次掷到 b≤tb\le t 就重掷，否则保留。则

E[B]=120(∑b=1t3.5+∑b=t+120b)=−t2+20t+42040,\mathbb{E}[B]=\frac{1}{20}\left(\sum_{b=1}^{t}3.5+\sum_{b=t+1}^{20}b\right)=\frac{-t^2+20t+420}{40},

该二次函数在 t=10t=10 处取最大。

因此 Bob 最优策略：第一次为 1..10 重掷，11..20 保留，此时 E[B]=13\mathbb{E}[B]=13。

所以

P(A>B)=1−1330=1730.P(A>B)=1-\frac{13}{30}=\boxed{\frac{17}{30}}.