囚犯假释

The Prisoner's Dilemma

专题: Brainteaser / 脑筋急转弯
难度: L4
来源: QuantQuestion

个人笔记

题目详情

量化面试题：囚犯假释：问“另一个会被释放的人”会改变概率吗。

英文原题

Three prisoners, $A,B$ , and $C$ , with apparently equally good records have applied for parole. The parole board has decided to release two of the three, and the prisoners know this but not which two. A warder friend of prisoner $A$ knows who are to be released. Prisoner $A$ realizes that it would be unethical to ask the warder if he, $A$ , is to be released, but thinks of asking for the name of one prisoner other than himself who is to be released. He thinks that before he asks, his chances of release are $\frac{2}{3}$ . He thinks that if the warder says " $B$ will be released," his own chances have now gone down to $\frac{1}{2}$ , because either $A$ and $B$ or $B$ and $C$ are to be released. And so $A$ decides not to reduce his chances by asking. However, $A$ is mistaken in his calculations. Explain.

解析

A 的计算错在：他忽略了“看守说了谁会被释放”本身也是随机事件，导致样本空间不对。

关键是：当释放组合为 $BC$ 时，看守在回答“B 会被释放”与“C 会被释放”之间需要某种规则（常见假设为等概率随机说一个）。

假设三种释放组合 $AB,AC,BC$ 先验各为 $1/3$ ，并且在 $BC$ 情况下看守以 $1/2$ 概率说 B、 $1/2$ 概率说 C。

若你听到看守说“B 会被释放”，则可能情形只有：

实际释放 $AB$ 且说 B：概率 $1/3$ ；
实际释放 $BC$ 且说 B：概率 $1/3\times 1/2=1/6$ 。

因此

\mathbb{P}(A\text{被释放}\mid \text{说B})=\frac{1/3}{1/3+1/6}=\boxed{\frac{2}{3}}.

所以 A 的释放概率并没有降到 1/2。

英文解析

Of all the problems people write me about, this one brings in the most letters.

The trouble with $A$ 's argument is that he has not listed the possible events properly. In technical jargon he does not have the correct sample space. He thinks his experiment has three possible outcomes: the released pairs $AB,AC,BC$ with equal probabilities of $\frac{1}{3}$ . From his point of view, that is the correct sample space for the experiment conducted by the parole board given that they are to release two of the three. But $A$ 's own experiment adds an event- the response of the warder. The outcomes of his proposed experiment and reasonable probabilities for them are:

$A$ and $B$ released and warder says $B$ , probability $\frac{1}{3}$ .
$A$ and $C$ released and warder says $C$ , probability $\frac{1}{3}$ .
$B$ and $C$ released and warder says $B$ , probability $\frac{1}{6}$ .
$B$ and $C$ released and warder says $C$ , probability $\frac{1}{6}$ .

If, in response to $A$ 's question, the warder says " $B$ will be released," then the probability for $A$ 's release is the probability from outcome 1 divided by the sum of the probabilities from outcomes 1 and 3 . Thus the final probability of $A$ 's release is $\frac{1}{3}\Big / \Big(\frac{1}{3} +\frac{1}{6}\Big)$ , or $\frac{2}{3}$ , and mathematics comes round to common sense after all.