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双骰收益：出现双 6 赢 100，出现单 6 亏 x

Double Dice Payoff

专题
Probability / 概率
难度
L2

题目详情

掷两枚公平六面骰：

  • 若出现双 6，赢 100。
  • 若出现恰好一个 6（另一个不是 6），亏损 xx。
  • 若两枚都不是 6，则继续重掷，直到出现前两种情况之一为止。

问：使得该游戏期望收益非负的最大 xx 是多少？

You roll two fair dice. If you get double 66s, you receive 100 dollar. If you get a 66 and a non-66, you lose xx. If you get anything else, you reroll both dice until you get double 66s or a 66 and non-66. What is the maximum value of xx where the game still has non-negative expected value?

解析

每轮：双 6 概率 1/361/36；恰好一个 6 的概率 10/3610/36；其余情况继续。

条件在“游戏终止”上归一化后：

P(双 6)=111,P(单 6)=1011.P(\text{双 6})=\frac{1}{11},\quad P(\text{单 6})=\frac{10}{11}.

期望收益为

100⋅111−x⋅1011≥0⇒x≤10.100\cdot\frac{1}{11}-x\cdot\frac{10}{11} \ge 0\Rightarrow x\le 10.

最大 x=10x=10。

Original Explanation

First we need to find the probability of every outcome. Double 66s occur with probability (16)2=136\left(\frac{1}{6}\right)^2 = \frac{1}{36}. We can count the number of 66 and non-66 combos there are and you’ll see there are 1010 (11-55 on the first die and 66 on the other with two different orderings). This gives that event a probability of 1036\frac{10}{36}. Since the other event is just a recurrence of these two events, we can normalize the probabilities of the two events we care about. We do this by finding the probability of our target event and divide it with the probability of the non-recurrent events. So the probability of double 66s in this case can be updated to 136(136+1036)=111\frac{\frac{1}{36}}{(\frac{1}{36} + \frac{10}{36})} = \frac{1}{11}. This gives the probability the game ends in a 66 and non 66 of 1011\frac{10}{11}. Finally, we can solve for xx. The profit equation here is 100⋅111−x⋅1011100 \cdot \frac{1}{11} - x \cdot \frac{10}{11} which has to be greater than or equal to 00. Thus 10011≥10x11\frac{100}{11} \geq \frac{10x}{11}. The maximum value of xx is thus 1010.