骰子游戏 3

Dice Game 3

专题: Probability / 概率
难度: L3
来源: OpenQuant

题目详情

你将获得一个游戏，你可以掷 2 个公平的 6 面骰子，并将总收入添加到你的总收入中。你可以滚动任意多次，但是，如果两个骰子都落在同一张面上，游戏就会停止，你会失去在此之前获得的一切。

你应该重新掷骰子以获得什么值？

You are offered a game where you roll 2 fair 6-sided die and add the sum to your total earnings. You can roll as many times as you'd like however, in the case where both die land on the same face, the games stops and you lose everything you gained until that point.

For what values should you re-roll?

解析

我们的决定归结为是否值得重新滚动？直观上，我们当前的金额越大，我们就越不应该基于损失大量的风险而重新滚动。只有当重新滚动的价值大于无风险兑现当前收益时，我们才应该重新滚动。

我们将当前收益称为 $x$ 。鉴于我们已经积累了 $x$ ，我们对重掷的预期值为 $\frac16(0) + \frac56(x+7)$ 。这是因为我们将以 $\frac16$ 的概率滚动相同的面，并以 $\frac56$ 的概率添加到我们的总和中。在我们添加总和的情况下，我们期望总和为 $7$ ，因为我们期望每个芯片的值为 $3.5$ 。

边际价值重新滚动应该大于我们的收入无风险，因此利用这一点我们可以形成我们的不平等： $\frac16(0) + \frac56(x+7) > x$ $\rightarrow x < \boxed{35}$ $35$ 是无差异点，因此我们应该滚动它之前的每个值并保留其之上的所有值。

Original Explanation

Our decision comes down to is it worth re-rolling? Intuitively, the greater our current sum, the less we should want to re-roll based on the risk of losing a lot. We should only re-roll when the value of re-rolling is greater than cashing out our current earnings risk free.

Let's call our current earnings $x$ . Our expected value on a re-roll given that we have already accumulated $x$ is $\frac16(0) + \frac56(x+7)$ . This is because we will roll identical faces with probability $\frac16$ and add to our sum with probability $\frac56$ . In the case we add to our sum, we are expecting a sum of $7$ as we expect a value of $3.5$ from each die.

The marginal value re-rolling should be greater than taking our earnings risk free so using this we can form our inequality: $\frac16(0) + \frac56(x+7) > x$ $\rightarrow x < \boxed{35}$

$35$ is the indifference point, thus we should roll for every value before it and keep all values above it.