I recently showed the famous Monty Hall problem to a friend. This friend solved the problem right away, and we realized quickly that the standard presentation of the problem is highly misleading.

Here’s the setup as it was originally described in the magazine column that made it famous:

Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?

I encourage you to think through this problem for yourself and come to an answer. Will provide some blank space so that you don’t accidentally read ahead.

Now, the writer of the column was Marilyn vos Savant, famous for having an impossible IQ of 228 according to an interpretation of a test that violated “almost every rule imaginable concerning the meaning of IQs” (psychologist Alan Kaufman). In her response to the problem, she declared that switching gives you a 2/3 chance of winning the car, as opposed to a 1/3 chance for staying. She argued by analogy:

Yes; you should switch. The first door has a 1/3 chance of winning, but the second door has a 2/3 chance. Here’s a good way to visualize what happened. Suppose there are a million doors, and you pick door #1. Then the host, who knows what’s behind the doors and will always avoid the one with the prize, opens them all except door #777,777. You’d switch to that door pretty fast, wouldn’t you?

Notice that this answer contains a crucial detail that is not contained in the statement of the problem! Namely, the answer adds the stipulation that the host “knows what’s behind the doors and will always avoid the one with the prize.”

The original statement of the problem in no way implies this general statement about the host’s behavior. All you are justified to assume in an initial reading of the problem are the observational facts that (1) the host happened to open door No. 3, and (2) this door happened to contain a goat.

When nearly a thousand PhDs wrote in to the magazine explaining that her answer was wrong, she gave further arguments that failed to reference the crucial point; that her answer was only true given additional unstated assumptions.

My original answer is correct. But first, let me explain why your answer is wrong. The winning odds of 1/3 on the first choice can’t go up to 1/2 just because the host opens a losing door. To illustrate this, let’s say we play a shell game. You look away, and I put a pea under one of three shells. Then I ask you to put your finger on a shell. The odds that your choice contains a pea are 1/3, agreed? Then I simply lift up an empty shell from the remaining other two. As I can (and will) do this regardless of what you’ve chosen, we’ve learned nothing to allow us to revise the odds on the shell under your finger.

Notice that this argument is literally just a restatement of the original problem. If one didn’t buy the conclusion initially, restating it in terms of peas and shells is unlikely to do the trick!

This problem was made even more famous by this scene in the movie “21”, in which the protagonist demonstrates his brilliance by coming to the same conclusion as vos Savant. While the problem is stated slightly better in this scene, enough ambiguity still exists that the proper response should be that the problem is underspecified, or perhaps a set of different answers for different sets of auxiliary assumptions.

The wiki page on this ‘paradox’ describes it as a veridical paradox, “because the correct choice (that one should switch doors) is so counterintuitive it can seem absurd, but is nevertheless demonstrably true.”

Later on the page, we see the following:

In her book The Power of Logical Thinking, vos Savant (1996, p. 15) quotes cognitive psychologist Massimo Piattelli-Palmarini as saying that “no other statistical puzzle comes so close to fooling all the people all the time,” and “even Nobel physicists systematically give the wrong answer, and that they insist on it, and they are ready to berate in print those who propose the right answer.”

There’s something to be said about adequacy reasoning here; when thousands of PhDs and some of the most brilliant mathematicians in the world are making the same point, perhaps we are too quick to write it off as “Wow, look at the strength of this cognitive bias! Thank goodness I’m bright enough to see past it.”

In fact, the source of all of the confusion is fairly easy to understand, and I can demonstrate it in a few lines.

Solution to the problem as presented

Initially, all three doors are equally likely to contain the car.
So Pr(1) = Pr(2) = Pr(3) = ⅓

We are interested in how these probabilities update upon the observation that 3 does not contain the car.
Pr(1 | ~3) = Pr(1)・Pr(~3 | 1) / Pr(~3)
= (⅓ ・1) / ⅔ = ½

By the same argument,
Pr(2 | ~3) = ½

Voila. There’s the simple solution to the problem as it is presented, with no additional presumptions about the host’s behavior. Accepting this argument requires only accepting three premises:

(1) Initially all doors are equally likely to be hiding the car.

(2) Bayes’ rule.

(3) There is only one car.

(3) implies that Pr(the car is not behind a door | the car is behind a different door) = 100%, which we use when we replace Pr(~3 | 1) with 1.

The answer we get is perfectly obvious; in the end all you know is that the car is either in door 1 or door 2, and that you picked door 1 initially. Since which door you initially picked has nothing to do with which door the car was behind, and the host’s decision gives you no information favoring door 1 over door 2, the probabilities should be evenly split between the two.

It is also the answer that all the PhDs gave.

Now, why does taking into account the host’s decision process change things? Simply because the host’s decision is now contingent on your decision, as well as the actual location of the car. Given that you initially opened door 1, the host is guaranteed to not open door 1 for you, and is also guaranteed to not open up a door hiding the car.

Solution with specified host behavior

Initially, all three doors are equally likely to contain the car.
So Pr(1) = Pr(2) = Pr(3) = ⅓

We update these probabilities upon the observation that 3 does not contain the car, using the likelihood formulation of Bayes’ rule.

Pr(1 | open 3) / Pr(2 | open 3)
= Pr(1) / Pr(2)・Pr(open 3 | 1) / Pr(open 3 | 2)
= ⅓ / ⅓・½ / 1 = ½

So Pr(1 | open 3) = ⅓ and Pr(2 | open 3) = ⅔

Pr(open 3 | 2) = 1, because the host has no choice of which door to open if you have selected door 1 and the car is behind door 2.

Pr(open 3 | 1) = ½, because the host has a choice of either opening 2 or 3.

In fact, it’s worth pointing out that this requires another behavioral assumption about the host that is nowhere stated in the original post, or Savant’s solution. This is that if there is a choice about which of two doors to open, the host will pick randomly.

This assumption is again not obviously correct from the outset; perhaps the host chooses the larger of the two door numbers in such cases, or the one closer to themselves, or the one or the smaller number with 25% probability. There are an infinity of possible strategies the host could be using, and this particular strategy must be explicitly stipulated to get the answer that Wiki proclaims to be correct.

It’s also worth pointing out that once these additional assumptions are made explicit, the ⅓ answer is fairly obvious and not much of a paradox. If you know that the host is guaranteed to choose a door with a goat behind it, and not one with a car, then of course their decision about which door to open gives you information. It gives you information because it would have been less likely in the world where the car was under door 1 than in the world where the car was under door 2.

In terms of causal diagrams, the second formulation of the Monty Hall problem makes your initial choice of door and the location of the car dependent upon one another. There is a path of causal dependency that goes forwards from your decision to the host’s decision, which is conditioned upon, and then backward from the host’s decision to which door the car is behind.

Any unintuitiveness in this version of the Monty Hall problem is ultimately due to the unintuitiveness of the effects of conditioning on a common effect of two variables.

In summary, there is no paradox behind the Monty Hall problem, because there is no single Monty Hall problem. There are two different problems, each containing different assumptions, and each with different answers. The answers to each problem are fairly clear after a little thought, and the only appearance of a paradox comes from apparent disagreements between individuals that are actually just talking about different problems. There is no great surprise when ambiguous wording turns out multiple plausible solutions, it’s just surprising that so many people see something deeper than mere ambiguity here.

## One thought on “The Monty Hall non-paradox”

1. sj2mi says:

There’s an interesting discussion of the MH problem in Gigerenzer’s Risk Savvy. He points out that in the real game show, you need to know if Monty Hall always offers contestants the option to switch. If not, you can be selected against. MH could only offer you the chance to switch when he can see you have selected the car. I like that “real world” addition to the problem. It is more akin to the types of problems insurers and other risk takers face in a world with asymmetric information. He comments that Monty “recalled that he rarely offered the switching option.”