The definitive Monty Hall explainer

2022 April 06

Imagine you’re on a gameshow. There are three doors, behind one of which is a car–” “You’re telling it wrong. There are three doors, behind one of which is a car.”

I have been obsessed for far too many years with incorrect statements of the Monty Hall problem, once going so far as to pick up a Korean pop-science book just to see if its explanation was wrong (conclusion: I can’t read Korean), and I fear the only way to sate this obsession is to write my own. Er, my own correct explanation, that is.

I will spoil the ending by giving the take-away lesson up front. The Monty Hall problem involves probability, which is to say it has an element of randomness. Randomness is a property of a process that yields some result, not a property of the result itself:

Source: https://xkcd.com/221/

Therefore any presentation of the Monty Hall problem that simply describes the sequence of events is incomplete, and has no single right answer: a complete statement must also give the underlying random process that yields the observed events.

Treating randomness as a process, not a result, is often the domain of computer science, but this is a broadly general concept that is vital to successfully navigating any confusing problem involving probability. Losing track of your sources of randomness and how they determine the observed outcomes is a common cause of errors. Perhaps related is one of the benefits of pre-committing to the statistical analysis performed in a scientific study – by designing your analysis before seeing the data, you are forced to grapple with the data generation process, rather than the particular observed data values.

The Monty Hall problem

The common presentation of the Monty Hall problem, as seen in any number of popular media, is akin to the following:

You are on a gameshow hosted by Monty Hall featuring 3 doors, one of which conceals a fancy car you want, and the other two hiding worthless gag “prizes”. The host asks you to choose which door’s prize you want: you choose door 1. He then opens door 3, revealing a gag prize, and gives you a chance to change your mind about which door to take. What is the probability that door 2 conceals the car?

The classic answers are either 1/2 or 2/3, with the latter being “correct”. However if you read the spoiler above, you already see a major issue: how can the answer be a probability between 0 and 1 if no source of randomness is stated in the problem?

It is commonly understood from the context of a gameshow that the three prizes are initially distributed behind the three doors uniformly at random, which serves as a source of randomness. With some re-wording we can suppose the player’s initial door choice was also at random, although this turns out to be redundant with the prizes being distributed at random. (Just as in rock-paper-scissors it is redundant to choose at random if you know your opponent is choosing at random.)

So let us consider the three possibilities:

  1. Case 1. Car is behind door 1, and events transpire as described.
  2. Case 2. Car is behind door 2, and events transpire as described.
  3. Case 3. Car is behind door 3, which is inconsistent with the specified sequence of events, so something else happens.

Again we are stuck: the problem as stated doesn’t tell us what happens in the 1/3 of cases that the car was behind door 3. Clearly something happens, but what?

The host’s behavior

What we are missing is a complete description of the host’s behavior. It is not enough to know what the host did in actuality, but also what the host would do in every counterfactual. We aren’t even told if the host’s behavior is random or deterministic! Indeed we could justify any answer from 0 to 1 by carefully choosing what procedure for the host to follow.

Suppose for example that the host’s procedure is to always open door 3. Then the cases are:

  1. Car is behind door 1 and host reveals door 3 has a gag prize.
  2. Car is behind door 2 and host reveals door 3 has a gag prize.
  3. Car is behind door 3 and host reveals door 3 has the car.

Then, the conditional probability that the car is behind door 2 given that the host reveals a gag prize is equal to 1/2. (Of course, the probability given that the host reveals a car is zero, since we can see the car is behind door 3.)

Note that, in this case, we don’t actually need to know how the player chooses which door for their initial selection, since it had no influence on the sequence of events. (Assuming that the player has no prior information about the location of the car and therefore that conditioning on their selecting door 1 does not change the probability of where the car is.)

Now suppose that the player’s procedure is to always select door 1, and the host’s procedure is to always open the lowest numbered door that is not the selection and is a gag prize. Then the cases are:

  1. Car is behind door 1 and host reveals door 2 has a gag prize.
  2. Car is behind door 2 and host reveals door 3 has a gag prize.
  3. Car is behind door 3 and host reveals door 2 has a gag prize.

Now, the conditional probability that the car is behind door 2 given that the host revealed a gag prize behind door 3 is 100%!

I was pleasantly surprised to see that the importance of specifying the host’s procedure was clearly explained by none other than Monty Hall himself, in a 1991 interview with the New York Times, a very rare example of math or science being accurately reported on in mainstream media:

[Monty Hall] picked up a copy of Ms. vos Savant’s original column, read it carefully, saw a loophole and then suggested more trials.

On the first, the contestant picked Door 1.

“That’s too bad,” Mr. Hall said, opening Door 1. “You’ve won a goat.”

“But you didn’t open another door yet or give me a chance to switch.”

“Where does it say I have to let you switch every time? I’m the master of the show. Here, try it again.”

[…] Whenever the contestant began with the wrong door, Mr. Hall promptly opened it and awarded the goat; whenever the contestant started out with the right door, Mr. Hall allowed him to switch doors and get another goat. The only way to win a car would have been to disregard Ms. vos Savant’s advice and stick with the original door.

Furthermore:

Dr. Diaconis and Mr. Gardner both noticed the same loophole when they compared Ms. vos Savant’s wording of the problem with the versions they had analyzed in their articles.

“The problem is not well-formed,” Mr. Gardner said, “unless it makes clear that the host must always open an empty door and offer the switch.”

This is not an idle objection. Any problem statement necessarily depends on the reader sharing some context to understand the meaning of the words being used, just as we inferred that the prizes were distributed randomly (and that the player’s initial selection is independent of where the prize is) from the context of it being a gameshow. However, it is a much bigger step to infer the host’s behavior from the incomplete information as given above. The original gameshow Let’s Make a Deal that Monty Hall hosted featured a wide assortment of variety games that did not follow any fixed format or rigid rules; it would have been entirely in keeping with that show for Hall to adapt the format dynamically with circumstances, or even entice players away from the main prize with cash incentives. So it is not reasonable to leave the reader to guess at the procedure Hall is following from contextual clues alone: the procedure should be fully specified.

Full statement of problem and solution

Here is a (more) complete statement of the Monty Hall problem:

A gameshow has three doors, one of which is chosen in advance uniformly at random to conceal a car; the others conceal gag prizes. A player is asked to select a door, and does so at random. Regardless of the choices so far, the host then opens a door, chosen at random from among those doors which the player did not select and does not have the car. Given that the player selected door 1 and the host opened door 3, what is the probability that door 2 has the car?

There are three sources of randomness (which are implied to be independent), and we need 12 cases to fully work out every possibility. However it is fairly clear that the problem is symmetric with respect to which door the player selected, so we can fix that as nonrandom without changing the problem. This leaves us with only four cases:

(It is simple enough to see how to write out the other 8 cases to allow for if the player selects door 2 or 3, remembering to divide the probabilities by 3 so they still add up to 1.)

In cases 2 and 3, the host is “randomly” choosing which door to open from only one possibility. The host’s randomness only is relevant in cases 1a and 1b.

The conditional probability is

\frac {P(\text{car behind 2 and host opens 3})}{P(\text{host opens 3})} = \frac {1/3}{1/6 + 1/3} = \frac 2 3.

We’ve used the counterfactual outcomes to assess the conditional probability of the outcome that occurred. If we can eliminate case 1a, for example, case 1b raises in probability to 1/3 and the answer we get at the end is 1/2. Alternatively, if we can add cases 2a, 2b, 3a, 3b in which the host reveals the car, then the probability of that branch lowers from 1/3 to 1/6 and we again get 1/2 as the answer.

If the incomplete Monty Hall problem, as typically stated, lacks sufficient information to uniquely determine the answer, what makes this version of the problem the correct way of completely specifying it? While correct statements of the Monty Hall problem are rare in pop science discussions of it, most agree on a more-or-less acceptable explanation of how to get to the answer 2/3. Working backwards from this solution one finds the problem being solved. Indeed, this was the original justification vos Savant gave when critics of her column which popularized the problem pointed out the ambiguities: she said that from her explanation of the answer it was clear what was intended in the problem. (And, in vos Savant’s defense, the original incomplete statement of the problem was not hers but quoting from an inquiring reader.)

Other sources and variants

While most popular discussions of the Monty Hall problem fail to properly state it, some do. Wikipedia, of course, gives the correct explanation, saying “The given probabilities depend on specific assumptions about how the host and contestant choose their doors.”, and has the most thorough discussion of the problem and its variants of any source I’ve seen. The New York Times interview with Hall, discussed above, doesn’t give an explicit mathematical treatment but lucidly captures the key point that the host’s behavior must be specified.

Three decades before vos Savant’s column popularized the Monty Hall problem, an exactly equivalent problem was given by Martin Gardner (the same who was quoted above):

Three prisoners, A, B, and C, are in separate cells and sentenced to death. The governor has selected one of them at random to be pardoned. The warden knows which one is pardoned, but is not allowed to tell. Prisoner A begs the warden to let him know the identity of one of the two who are going to be executed. “If B is to be pardoned, give me C’s name. If C is to be pardoned, give me B’s name. And if I’m to be pardoned, secretly flip a coin to decide whether to name B or C.”

The warden tells A that B is to be executed. Prisoner A is pleased because he believes that his probability of surviving has gone up from 1/3 to 1/2, as it is now between him and C. Prisoner A secretly tells C the news, who reasons that A’s chance of being pardoned is unchanged at 1/3, but he is pleased because his own chance has gone up to 2/3. Which prisoner is correct?

Being both unambiguous and having priority in time, I would prefer if this formulation supplanted the classic presentation of the Monty Hall problem, but of course its very ambiguity led to its popularity.

A popular variant is to suppose there are 100 doors, of which the host opens 98. (vos Savant proposed 1000000 doors, but I guess people got tired of writing that many zeros.) This improves the intuition behind the correct solution to the problem but doesn’t by itself improve the clarity of the problem statement. Similarly, consider another formulation:

Three tennis players. Two are equally-matched amateurs; the third is a pro who will beat either of the amateurs, always.

You blindly guess that Player A is the pro; the other two then play. Player B beats Player C. Do you want to stick with Player A in a Player A vs. Player B match-up, or do you want to switch? And what’s the probability that Player A will beat Player B in this match-up?

After writing this article I found an interesting twist in which the player bribes the host in advance to choose their behavior, subject to the limitation that the host must open exactly one door that is not the player’s selection nor the car. The goal then is to choose a strategy (for both player and host) to maximize the chance of getting the car, and calculate that probability.

(Note that some of the behaviors I’ve mentioned above result in getting the car with 0% or 100% probability, but they do not satisfy the specified constraint on the host’s behavior, so we can’t use them here.)

Suppose, say, the bribed host opens door 3. This outcome occurs in every case that the car is behind door 2 (i.e. “case 2” above), and between 0% and 100% of the cases that the car is behind door 1 (i.e. “case 1b” above), with the probability depending on what the player bribed the host to do. Thus here the player does at least as well by switching as by staying, and likewise if the bribed host opened door 2 (cases 3 and 1a above). Therefore always switching is an optimal strategy for the player, with which the player wins in cases 2 and 3, with a probability of 2/3 – regardless of the behavior of the host. (Though for certain host behaviors this is not the only optimal strategy.)

The only thing the player can change by bribing the host is the distribution of wins and losses between doors 2 and 3: e.g., so that whenever the host opens door 3 it is a guaranteed win, but at the cost that they only win half the time that door 2 is opened. The overall win rate will still be the same.

Most popular presentations of the Monty Hall problem ask only whether you should switch, but as we just saw, switching dominates staying in most interpretations of the problem, side-stepping the central issue; this is why I phrased the problem as asking more specifically for the probability of finding the car.

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