# Understanding the Monty Hall Problem

I have extensively revised, expanded, and republished this post as “The Compleat Monty Hall Problem.” Please go there.

I first encountered  the Monty Hall problem a few years ago. My intuitive answer to the problem was wrong, according to the explanation of the problem that I read at the time. I forgot about the problem until a few days ago, when I gave the answer I had given before and read, again, that I had answered incorrectly. With two strikes, I began to focus on the problem, in an effort to understand why my intuitive answer is incorrect, and why the “correct”  answer is indeed correct.

What is the Monty Hall problem? It’s a probability puzzle named for the original host of an old TV game-show, Let’s Make a Deal. The puzzle acquired its name because the setup of the puzzle is similar to one of the challenges faced by contestants who appeared on the show. The puzzle goes like this:

1. There are three doors: 1, 2, and 3.

2. Behind one of the doors is an item of some value, which I denote with \$\$.

3. Behind each of the other two doors are items of little or no value, which I denote with ε (epsilon, often used in scientific notation to represent an error term).

4. The contestant chooses one of the doors, but the item behind it is not revealed (yet).

5. The host, who knows what is behind each of the doors, opens a door other than the one chosen by the contestant. The host always opens a door to reveal an ε. He is able to do so because no matter which door the contestant chooses, there is at least one other door that conceals an ε.

6. The host then asks the contestant if he wants to stick with the door he has chosen and accept the prize hidden behind it, or if he wants to switch to the other unopened door and accept the prize hidden behind that door. The host asks the question regardless of what is behind the door that the contestant chose originally; the question is not asked for the purpose of enticing the contestant to abandon a choice that would yield \$\$.

This is where the intuitive answer fails. The intuitive answer is to stick with the door already chosen. Why? It seems that the usual assumption is that there is an equal probability of finding \$\$ behind any door, so that switching won’t necessarily lead to a better outcome. Nor (if the usual assumption is correct) would switching lead to a worse outcome, so the contestant might as well switch. But there seems to be a psychological preference for standing pat.

Unlike many situations in which snap judgments are necessary and often effective — because they are based on training, practice, and the ability to pick up subtle clues (e.g., the spin on a pitched ball, the moment at which an opponent is vulnerable to counterattack) — the usual snap judgement about the Monty Hall problem is wrong. It is wrong because the assumption about probabilities is wrong. This becomes obvious when the problem is modeled:

The usual intuitive answer focuses on one  possible outcome (A, for instance), to the exclusion of other  possible outcomes (B and C). And the usual intuitive answer would be correct if A were the only possible outcome. But A is not the only possible outcome. The contestant’s choice of Door 1 does not preclude B and C as possible outcomes.

Given that B and C are also possible outcomes, the probability that \$\$ is behind Door 1 is 1/3. And that probability remains unchanged by the host’s revelation.

Further, given that B and C are also possible outcomes, the contestant’s real choice is between Door 1 (in this example) and whichever door the host does not open — not between Door 1 and Door 2 or Door 1 and Door 3. A look at the figure above tells you that if the contestant chooses whichever door the host does not open (either Door 2 or Door 3), the probability of revealing \$\$ is 2/3 (two of the three possible outcomes). (See this video for an elegantly simple explanation of the same principle.)

In general: The array of possible outcomes (A, B, and C) is always the same (even if displayed in a different order), so it does not matter whether the contestant’s initial choice is Door 1, Door 2, or Door 3. The probabilities associated with the contestant’s initial and second choices remain unchanged. You can work it out for yourself.

The Monty Hall problem is the kind of problem that warrants careful thought instead of a snap judgment. The Monty Hall problem should be a lesson to anyone who is confronted with a situation in which a snap judgment is unnecessary. The snap judgment can lead you badly astray.