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无限次重掷但每次重掷付 1

Basic Die Game VI

专题
Probability / 概率
难度
L4

题目详情

Alice 掷一枚公平六面骰（点数 1..6），看到点数后可以选择停下并拿到该点数作为收益。

她也可以选择重掷，但每次重掷需要支付 1（从最终收益中扣除）。重掷次数不受限。

在最优策略下，期望收益是多少？

Alice rolls a fair 66-sided die with the values 1−61-6 on the sides. She sees that value showing up and then is allowed to decide whether or not she wants to roll again. Each re-roll costs $1. Whenever she decides to stop, Alice receives a payout equal to the upface of the last die she rolled. Note that there is no limit on how many times Alice can re-roll. Assuming optimal play by Alice, what is her expected payout on this game?

解析

最优策略为设置停的阈值（掷到至少某个点数就停，否则重掷）。

比较各阈值可得最优期望收益为 4。

Original Explanation

Let vkv_k be the value of the game if you accept a roll of kk or more and re-roll otherwise. Our optimal strategy will be in this form because we can view each round of the game as an independent trial. Therefore, our strategy should be the same between trials. Then

v1=(1/6)(1+2+3+4+5+6)⇒v1=7/2v2=(1/6)(v2−1)+(1/6)(2+3+4+5+6)⇒v2=19/5v3=(2/6)(v3−1)+(1/6)(3+4+5+6)⇒v3=4v4=(3/6)(v4−1)+(1/6)(4+5+6)⇒v4=4v5=(4/6)(v5−1)+(1/6)(5+6)⇒v5=7/2v6=(5/6)(v6−1)+(1/6)(6)⇒v6=1\begin{array}{lll} v_1=(1 / 6)(1+2+3+4+5+6) & \Rightarrow & v_1=7 / 2\\ v_2=(1 / 6)\left(v_2-1\right)+(1 / 6)(2+3+4+5+6) & \Rightarrow & v_2=19 / 5\\ v_3=(2 / 6)\left(v_3-1\right)+(1 / 6)(3+4+5+6) & \Rightarrow & v_3=4 \\ v_4=(3 / 6)\left(v_4-1\right)+(1 / 6)(4+5+6) & \Rightarrow & v_4=4 \\ v_5=(4 / 6)\left(v_5-1\right)+(1 / 6)(5+6) & \Rightarrow & v_5=7 / 2 \\ v_6=(5 / 6)\left(v_6-1\right)+(1 / 6)(6) & \Rightarrow & v_6=1 \end{array}

All of these are derived from the fact that if we obtain a value at least our minimum stopping value, we just stop immediately. Otherwise, we just go again and it is the same game but we lose 11 in payout from the cost of the roll. Therefore, this means that our optimal EV is 44.