The Monty Hall Problem: Why You Should Always Switch

Three boxes sit on a table. One contains a cat. Two contain empty bowls. You choose one. The host, who knows which box holds the cat, opens one of the other boxes to reveal an empty bowl, then offers you the chance to switch.

Should you switch?

Yes. Switching wins the cat with probability 2/3. Staying wins with probability 1/3. This result has generated enough confident disagreement over the decades, including from professional mathematicians writing indignant letters, that it's worth going through carefully.

Why the 50/50 Intuition Is Wrong

The common objection: after the host opens a box, there are two remaining boxes, one of which holds the cat. So it's 50/50. The error in this reasoning is subtle but decisive.

The host does not open a box at random. The host knows where the cat is and always opens a box that contains a bowl. This restriction on the host's behavior encodes information into the action. When the host opens a box, they are not just eliminating one option. They are telling you something specific about the box they could not open.

If a host opened a box at random and happened to reveal a bowl, then the remaining probability would indeed be split evenly. But that is not the game. The host's knowledge is not random, and neither is their action.

Bayes' Theorem

Label the boxes 1, 2, and 3. You choose box 1. The host opens box 3, revealing a bowl. Should you switch to box 2?

Apply Bayes' theorem. The probability you want is $P(\text{cat in 2} \mid \text{host opens 3})$.

$$P(\text{cat in 2} \mid \text{host opens 3}) = \frac{P(\text{host opens 3} \mid \text{cat in 2}) \cdot P(\text{cat in 2})}{P(\text{host opens 3})}$$

Work through each term:

• $P(\text{cat in 2}) = 1/3$. All three boxes had equal prior probability.
• $P(\text{host opens 3} \mid \text{cat in 2}) = 1$. If the cat is in box 2, the host must open box 3, since box 1 is your chosen box and box 3 is the only option left.
• $P(\text{host opens 3}) = 1/2$. The host opens box 2 or box 3 with equal probability when the cat is in your chosen box (box 1); always opens box 3 when the cat is in box 2; never opens a cat box. Averaging over the three equally likely prior locations gives $P(\text{host opens 3}) = 1/2$.

Substituting:

$$P(\text{cat in 2} \mid \text{host opens 3}) = \frac{1 \cdot \frac{1}{3}}{\frac{1}{2}} = \frac{2}{3}$$

Staying with box 1 gives the complementary probability: $1/3$.

The Enumeration

If probability calculations feel abstract, enumerate. There are three equally likely initial configurations:

Cat in box 1: You chose correctly. The host opens box 2 or 3 (doesn't matter which). Switching loses.

Cat in box 2: You chose box 1. The host must open box 3 (box 2 has the cat, box 1 is yours). Switching wins.

Cat in box 3: You chose box 1. The host must open box 2 (box 3 has the cat, box 1 is yours). Switching wins.

Two of three initial configurations produce a win when you switch. One produces a win when you stay. The strategy of always switching wins 2/3 of the time.

Simulations confirm this. Run the experiment 10,000 times with a program that randomizes the cat's location and has the host follow the rules. Switching wins close to 6,667 times. Staying wins close to 3,333.

The Host Matters

The key insight is that the host's action carries information precisely because it is constrained. A host who opens a box uniformly at random, and happens to reveal a bowl, is giving you weaker evidence. In that case, the probability does update toward 50/50 for the remaining box.

But the Monty Hall host is not operating at random. They will never reveal the cat. That constraint means the box the host chose not to open has absorbed probability from the box they were forced away from.

Another way to see this: if there were 100 boxes and you chose one, and then the host opened 98 of the remaining 99 boxes to reveal empty bowls, you should switch with considerable confidence. The host's knowledge forced them to leave that one box alone. The probability that the cat is in it is 99/100. The Monty Hall problem is the same structure with less dramatic numbers.

Why Intuition Fails

Human probability intuition tends to treat independent-looking events as independent. After the host acts, there are two closed boxes, and the mind assigns them equal probability by default. The error is in treating the host's action as neutral. It is not neutral. The host could not open the winning box. That restriction is precisely what makes the remaining unchosen box carry the weight it does.

The confident wrongness this problem produces is instructive. It is a reminder that probability requires counting carefully, not reaching for the nearest symmetric-looking scenario.