Can you convincingly explain the Monty Hall problem?
Four approachs for an elusive puzzle
The Monty Hall problem is a well-known probability puzzle that can lead to some not-so-well-appreciated challenges.
It first rose to prominence in 1990, when Marilyn vos Savant published the puzzle in a magazine column. (If you’re not familiar with the problem, a recap follows.) Many readers, including university mathematicians, were sceptical of the solution she provided. One letter read: ‘Our math department had a good, self-righteous laugh at your expense.’ ‘May I suggest that you obtain and refer to a standard textbook on probability before you try to answer a question of this type again?’ wrote another.
Recap of the problem
The premise of the Monty Hall problem is simple. You’re on a game show. There are three doors. Behind one is a car; behind the other two, goats. You start by picking a door. The host, Monty Hall—who knows what’s behind each door—opens a different door, revealing a goat. He then offers you a choice: stick with your original pick or switch to the remaining unopened door.
What should you do?
The proof problem
It might seem like it makes no difference. Intuitively, you might assume the chance of winning is 50:50 with two doors left. But the correct answer is that you should switch. If you’ll do this, you’ll have a 2/3 chance of winning the car.
The Monty Hall problem is often recounted as an example of the counterintuitive nature of probability, which can lead even specialists astray. Some people initially get it wrong, goes the usual account, but there’s a correct answer, and that’s the end of it.
I first heard about the problem as a teenager, a decade after the vos Savant controversy. When I first encountered the puzzle, I was quickly convinced of the solution—and had assumed that was that. But, crucially, I didn’t realise how difficult it would be to convince others.
The Monty Hall problem makes a brief appearance in my new book Proof. Now I’ve started doing talks and podcasts to mark the UK and US launch, I’ve noticed I often get asked about the problem by readers. But it’s not just about the problem. Even when someone has previously heard both the problem and the solution, they are still not satisfied. For many, the Monty Hall problem causes lingering frustration.
So here is my attempt to run through some different ways of thinking about the problem—and how you might be able to convince yourself and others of the correct solution.
1. Proof by exhaustion
A common starting point when dealing with a decision-based problem is to run through all possible outcomes. Here’s a reminder of the set-up:
There are 3 initial choices (the car is randomly behind one of three doors).
Monty always opens a goat door (never the car).
You can either stay or switch.
This produces the following set of possibilities:
As you can see, in 2 out of 3 cases, switching wins the car.
2. Proof by simulation
If writing out the possibilities didn’t convince you, let’s create a simulation of the game with the following steps:
Randomly place the car.
Make a random initial pick.
Simulate Monty revealing a goat.
Track outcomes for both stay and switch strategies.
The following graph shows results from 1,000 simulated games. We see that across multiple games, the switching strategy consistently performs better, winning around 2/3 of the time on average:
3. Using an extreme example
If writing down possibilities and running simulations still isn’t fully convincing, we can instead use a more extreme example as a thought experiment. Suppose there are 100 doors, rather than just 3. You pick one of these 100 doors. Then Monty opens 98 others, all goats.
Your original pick had a 1/100 chance of being correct. That remains the case. Hence one other unopened door now has a 99/100 chance of hiding the car.
The same logic applies with three doors—it’s just harder to see initially, because the difference in odds isn’t as stark.
4. Grouping proof
A related way to approach the problem is to frame it in terms of two groups: your choice and not your choice.
Your single door has a 1/3 chance of being the winner. The group of the two other doors therefore has a 2/3 chance. Monty’s reveal of the goat simply tells you where the car isn’t within that group. So there’s still a 2/3 probability for the ‘other doors’ group, but only one door left that you can open.
By switching, you can claim the full odds of the larger group.
From rote solutions to proof
Even if you’ve heard of the Monty Hall problem before—and remembered the correct answer when you saw the doors—how confident are you that you’d be able to give others a satisfactory explanation?
If put on the spot, could you move beyond a memorised answer and into the deeper terrain of proof? And which of the above routes would you choose?
Arguably the real puzzle isn’t solving the Monty Hall problem—but working out how to explain it clearly to someone else.
If you’re interested in reading more about the Monty Hall problem and the perils of proving things, you might like my new book Proof: The Uncertain Science of Certainty, which is available now.
Hi, Adam,
As a mathematician, I've long followed the MH problem. I basically used the exhaustive approach when explaining it to others, but your "grouping" approach is far better--a perfect conceptual shortcut. Thanks!
I think it's relevant that you're not trying to convince a naive person who's a total blank slate with no idea what the answer might be — good luck finding someone like that! — but rather, you want to convince someone who has good reason to think the answer is other than it is.
So while it's definitely helpful to have multiple explanations for why the right answer is right, I think you'll have more success if you also explain why the intuition goes awry, so that the person can successfully set their intuition aside and be open to a different answer.
To that end, I'd suggest something like:
> The obvious answer is "It doesn't matter": each door originally had equal probability of hiding the car, and Monty Hall hasn't moved the car, so they still have equal probability.
> But by that argument, you could also say that it's just as good to switch to the door where Monty Hall just revealed a goat! After all, he didn't change whether it had hidden a car, so surely it still has a one-third chance of hiding a car, even though you can plainly see that it doesn't?
> The resolution is to see that the car never literally had a one-third chance of being behind each door — it was behind a specific door, so that door had a 100% chance and the others had a 0% chance — it's just that you had no information about which door it was behind, so from your perspective it was *as if* each door had the same chance. If you played this game many times, your first guess would be right about one-third of the time.
> Monty Hall has now changed the situation: he's given you some information (that such-and-such door hid a goat), and crucially, he decided what information to give you *based on what door you had chosen*. By choosing one door, you guaranteed that Monty Hall wouldn't open it, so you affected what information he could give you. If you chose a door with a goat, then you forced him to show you the other door with a goat, whereas if you chose a door with a car, then you let him freely choose which goat to show you. So he's more likely to show you a given goat if you chose the door with the other goat than if you chose the door with the car; and now he's showing you a goat, so working backwards, this means it's more likely that you picked the other goat than that you picked the car.
> (Of course, the car is still behind one specific door, so it still technically has a 100% probability of being behind that door and a 0% probability of being behind the other. But given the information you now have, it's *as if* it has a one-third chance of being behind one door and a two-thirds chance of being behind the other. If you play this game many times using this strategy, you'll win about two-thirds of the cars.)