三门问题

若“观众随机开门，且这次恰好开到空门”，则在观察到“开到的是空门”这一条件下，换与不换 胜率相同。

设你初选门 3，观众在门 1 与门 2 中等概率随机选一扇打开。观察到他打开门 2 且为空。

记事件 $E$ 为“打开门 2 且为空”。

• 若奖在门 3（先验 $1/3$），则门 1、2 都空，$\mathbb{P}(E\mid奖=3)=1/2$。
• 若奖在门 1（先验 $1/3$），门 2 为空，$\mathbb{P}(E\mid奖=1)=1/2$。
• 若奖在门 2（先验 $1/3$），门 2 不空，$\mathbb{P}(E\mid奖=2)=0$。

所以后验

$$\mathbb{P}(奖=3\mid E)=\mathbb{P}(奖=1\mid E)=\frac{1}{2}.$$

因此换到门 1 的胜率为 $1/2$，不换留在门 3 的胜率也为 $1/2$，换不换都一样。

（若不加“恰好开到空门”的条件，而是允许观众可能开到奖门，则游戏机制已不同，需另行建模。）

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英文解析

There are two interpretations. If we assume that the game can be played repeatedly with an audience member always revealing a door to be empty, then we must also assume that the audience member knows the location of the prize. Otherwise, how can he or she always reveal an empty door in repeated play? In this case, if we make the same uniformly random assumptions about prize placement and empty doors revealed as in the previous question, then the audience member is, in effect, wearing the host's hat, and the argument is the same as in the previous question: You should always switch.

Suppose instead that the host of this week's show has, just for a one- off special occasion (his 60 th birthday, say), decided to let an audience member (who is ignorant of the prize's location) reveal another door. Suppose it just happens to turn out to be empty. We cannot now talk about a frequentist approach and repeated plays of the game because in repeated plays, the audience member would reveal the prize one third of the time, and that is not the situation we find ourselves in. The ignorant audience member has, on this occasion, just happened to show us an empty door. The audience member has blindly removed one door from the sample space, and we have a 50/50 chance of winning (and also of losing) if we switch. We are therefore indifferent.