找出钱箱

Find the Money Box

专题: Probability / 概率
难度: L3
来源: OpenQuant

题目详情

有五个盒子：其中一个装着 $150，其余都是空的。你可以在任意时刻支付 $X$ 美元打开一个盒子并拿走里面的东西。假设你会一直付费直到找到装钱的盒子，问 $X$ 取多少（美元）时这个游戏是公平的？

There are five boxes: one with $150, the rest are empty. At any time, you may pay $X$ dollars to open one of the boxes and keep the contents. Assuming you continue paying until you find the box with the money, what is the value $X$ (in dollars) that makes the game fair?

解析

由于无法提前知道哪个盒子有 $150，且每个盒子等可能，因此你第 $k$ 次打开就找到钱的概率为 $\frac{1}{5}$ （ $k=1,2,3,4,5$ ）。

设 $N$ 为打开的盒子数，则

E[N]=\frac{1+2+3+4+5}{5}=3.

游戏公平意味着期望成本等于期望收益：

E[\text{cost}]=XE[N]=150\ \Longrightarrow\ X\cdot 3=150\ \Longrightarrow\ \boxed{X=50}.

Original Explanation

Since there is no way to know in advance which box contains the $150, each box is equally likely to contain the money. The probability of finding the money in each box is $\frac{1}{5}$ .

The expected number of boxes opened before finding the $150 can be calculated using the concept of expectation for discrete random variables.

The probability of finding the money in the first box is $\frac{1}{5}$ , and in this case, only 1 box is opened.
The probability of finding the money in the second box is $\frac{1}{5}$ , and in this case, 2 boxes are opened.
The probability of finding the money in the third box is $\frac{1}{5}$ , and 3 boxes are opened, and so on.

The expected number of boxes opened, $E[boxes]$ , is the weighted sum of these possibilities

E[boxes] = (1 \times \frac{1}{5}) + (2 \times \frac{1}{5}) + (3 \times \frac{1}{5}) + (4 \times \frac{1}{5}) + (5 \times \frac{1}{5})

This simplifies to:

E[boxes] = \frac{1 + 2 + 3 + 4 + 5}{5} = \frac{15}{5} = 3

Thus, on average, 3 boxes will be opened before finding the one with $150.

For the game to be fair, the expected cost should equal the expected reward. The reward for this game is $150, so the fair game condition is:

\text{Total expected cost} = \text{Expected reward} \\ 3X = 150 \\ \boxed{X = 50}