两信封悖论

Two Envelopes Paradox

专题: Probability / 概率
难度: L2
来源: QuantQuestion

题目详情

你面前有两个外观完全相同的信封，其中一个信封里的钱是另一个的两倍。

你随机选了一个信封并看到其中金额为 $X$ （但看不到另一个信封）。

问：为了最大化期望收益，你应该保留还是交换信封？

You are given two indistinguishable envelopes. One envelope contains twice as much money as the other. Without knowing the amounts, you choose one at random. After looking at the amount $X$ inside your chosen envelope (but not the other), you must decide:
Should you keep the money or switch envelopes to maximize expected gain?

解析

这个“总是该换”的推导是著名悖论，关键在于直接把“ $X$ 是较小/较大金额各 1/2”套进去会隐含错误的先验。

如果不额外假设金额的先验分布，仅凭看到 $X$ 不能得出“永远换更好”的结论。

因此：

在没有先验分布/上限信息时，该问题本身是悖论式的，不能仅靠简单期望计算得到正确策略。
若给定合理先验（例如金额有上界或分布已知），可根据后验计算决定是否换。

Original Explanation

Let the amount in your chosen envelope be $X$ .

There are two possibilities:

With probability $\frac{1}{2}$ , $X$ is the smaller amount, so the other envelope contains $2X$ .
With probability $\frac{1}{2}$ , $X$ is the larger amount, so the other envelope contains $\frac{X}{2}$ .

Now, calculate the expected value $E$ of the other envelope:

E = \frac{1}{2} \cdot 2X + \frac{1}{2} \cdot \frac{X}{2} = X + \frac{X}{4} = \frac{5}{4}X

So, the expected amount in the other envelope is greater than $X$ .

✅ Conclusion:
On average, the other envelope contains $\frac{5}{4}$ times the amount in your current envelope.
It seems you should always switch!