应该换门。

初选中奖概率为 $1/3$，初选没中概率为 $2/3$。

主持人知道奖品位置且必开一扇空门：

• 若你初选没中（概率 $2/3$），剩下那扇未开的门必是奖品门；
• 若你初选中了（概率 $1/3$），换门会换到空门。

因此换门中奖概率为 $\boxed{2/3}$，不换为 $\boxed{1/3}$。

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英文解析

The "Let's Make a Deal" or "Monty Hall" problem is very frequently asked. Many people find it very difficult.

Assume that you choose Door 3. The host opens Door 2 and offers you the chance to switch to Door 1. Should you do it? If you have decided that it does not matter whether you switch doors or not (indifference), or that you should definitely not switch (aversion), then you should go back and think again before reading any further. Stop here and try again.

Let me begin with very simple intuition. My experience, however, is that many readers cannot accept the simple intuition, and for them I provide a formal proof using Bayes' Theorem.

SMPPLE INTUITION

Assume for a moment that you have already decided that you will switch doors. What then is the probability that you will find the prize behind the door you switch to? Well, you win the prize if you originally chose one of the two doors that has nothing behind it. In that case, the host shows you the other empty door, and switching yields the prize. So, the problem reduces to figuring the probability that you originally

chose one of the two doors that has nothing behind it. That unconditional probability is just two thirds by construction. You thus have probability two thirds that you win by switching and one third that you lose by switching. So, you should switch!2

Formal Bayer's Theorem Proof

If you are to play this game repeatedly, two- thirds of the time you profit by switching, and one- third of the time you lose by switching. Let $B_{k}$ denote the event that the prize is behind Door number $k$ (" $B$ " for behind). Let $H_{j}$ denote the event that you see the host open Door number $j$ (" $H$ " for host).

The unconditional probabilities of the location of prizes (probabilities calculated without conditioning on which door the host opens) are simply $P\left(B_{1}\right) = P\left(B_{2}\right) = P\left(B_{3}\right) = \frac{1}{3}$ . What you need to know is the conditional probability $P\left(B_{1} \mid H_{2}\right)$ . That is, the probability that the prize is behind Door 1 given that you see (or "conditional on") the host open Door 2. We use a straightforward application of conditional expectations and Bayes' Theorem (see Feller [1968, Chapter V]), as follows:

$$P\left(B_{1} \mid H_{2}\right) = \frac{P\left(B_{1} \cap H_{2}\right)}{P\left(H_{2}\right)} = \frac{P\left(H_{2} \cap B_{1}\right)}{P\left(H_{2}\right)} = \frac{P\left(H_{2} \mid B_{1}\right) \times P\left(B_{1}\right)}{P\left(H_{2}\right)}$$

You know that $P\left(B_{1}\right) = \frac{1}{3}$ , but what about $P\left(H_{2} \mid B_{1}\right)$ and $P\left(H_{2}\right)$ ? You know that the host is going to show you an empty door other than the door you choose (assume through all of this that it is Door 3 that you choose). The host's door must be revealed empty and cannot be the same door that you choose. Therefore, it must be that if you choose Door 3, then $P\left(H_{2} \mid B_{1}\right) = 1$ .

Now, $P\left(H_{2}\right)$ is given by

$$\begin{array}{r l} & {P\left(H_{2}\right) = \left[P\left(H_{2}\mid B_{1}\right)\times P\left(B_{1}\right)\right] + \left[P\left(H_{2}\mid B_{2}\right)\times P\left(B_{2}\right)\right]}\\ & {\qquad +\left[P\left(H_{2}\mid B_{3}\right)\times P\left(B_{3}\right)\right],} \end{array}$$

so some extra terms need to be calculated to get $P\left(H_{2}\right)$ . Well, the host's door must be shown to be empty, so it must be that $P\left(H_{2} \mid B_{2}\right) = 0$ . The host is impartial, so it

must be that $P(H_{2} \mid B_{3}) = \frac{1}{2}$ [and $P(H_{1} \mid B_{3}) = \frac{1}{2}$ ]. Thus, $P(H_{2})$ is given by

$$\begin{array}{l}{P(H_{2}) = [P(H_{2}\mid B_{1})\times P(B_{1})] + [P(H_{2}\mid B_{2})\times P(B_{2})]}\\ {+[P(H_{2}\mid B_{3})\times P(B_{3})]}\\ {= \left(1\times \frac{1}{3}\right) + \left(0\times \frac{1}{3}\right) + \left(\frac{1}{2}\times \frac{1}{3}\right) = \frac{1}{2}.} \end{array}$$

It follows that the probability of finding the prize if you switch doors is twothirds:

$$P\left(H_{1} \mid H_{2}\right) = \frac{P\left(H_{2} \mid B_{1}\right) \times P\left(B_{1}\right)}{P\left(H_{2}\right)} = \frac{1 \times \frac{1}{3}}{\frac{1}{2}} = \frac{2}{3}$$

The summary in Table D.5 may clarify matters further. You choose Door 3. The host must choose an empty door to open. If the prize is behind Door 1, he must open Door $2\left[P\left(H_{2} \mid B_{1}\right) = 1\right]$ . However, if the prize is behind Door 3, he can choose between Doors 1 and $2\left[P\left(H_{2} \mid B_{3}\right) = \frac{1}{2}\right]$ . If you see Door 2, it is either because the prize is behind Door 1, and the host had no choice, or it is because the prize is behind Door 3, and the host randomly chose between Doors 1 and 2. It, therefore, follows that if you choose Door 3, and Door 2 is revealed empty by the host, the prize is twice as likely to be behind Door 1 as it is to be behind Door 3. Continuing along this line of thought, we may take

Table 4.1: The Monty Hall Problem

<table><tr><td colspan="4">Assume You Choose Door 3</td></tr><tr><td>Prize Location Bj</td><td>Host Opens Hi</td><td>Unconditional Probability P(Hi ∩ Bj)</td><td>Conditional Probability P(Hi | Bj)</td></tr><tr><td>1</td><td>2</td><td>1/3</td><td>1</td></tr><tr><td>2</td><td>1</td><td>1/3</td><td>1</td></tr><tr><td rowspan="2">3</td><td>1</td><td>1/6</td><td>1/2</td></tr><tr><td>2</td><td>1/6</td><td>1/2</td></tr></table>

a frequentist approach. Suppose you play the game repeatedly and always choose Door 3. If you look at all the times the host reveals Door 2 empty, you will find that two- thirds of the time the prize lies behind Door 1, and one- third of the time it is behind Door 3. Seeing Door 2 empty is thus a stronger signal that Door 1 has the prize than it is that Door 3 has it. This argument is more general, of course. Whichever door you choose, seeing the host reveal an empty door is a signal that you

should switch.