Why the Monty Hall problem is confounding
When I was in eighth grade, I became obsessed with a card game called Killer Bunnies, spending an embarrassing amount of time on a fan forum about its mechanics and strategy. I have no idea what we actually talked about regarding the game, but there's one thread I'll never forget: a heated, multi-month debate about the Monty Hall problem.
The Monty Hall problem is a brain teaser based on the 20th century game show Let's Make a Deal. Host Monty Hall shows you three doors; he tells you that one contains a car, and the other two each contain a goat. You pick one, and Monty says, let's build the suspense and show you what's behind one of the doors you didn't pick—which is inevitably a goat. Then he asks, do you want to switch to the other unopened door?
The answer is yes: switching doubles your chances of getting the car. (Assuming, of course, that you want the car more than the goat.)
If you've never seen this problem before, that sounds crazy; if you've seen it and understood the solution, it sounds obvious. That's true of most brain teasers, though. The Monty Hall problem is uniquely fascinating because everyone who knows the answer is absolutely terrible at explaining it to others.
I've heard dozens of people try to explain the solution, and not only has it never worked, you can tell right from the start that it's going nowhere. "You start with a 1/3 probability of being right, and that doesn't change," they usually start, which sounds like (and is) bad statistics. "This will help: imagine there are 1000 doors with 999 goats...", they continue, and that in fact never helps. They're in good company: the puzzle became famous when Marilyn vos Savant, then billed as the world's smartest person, gave the correct solution in a magazine column and got thousands of letters from irate PhDs arguing she was wrong.
(Mathematicians studying the multi-armed bandit problem during World War II supposedly claimed it was so frustrating that they should drop it on Germany so enemy scientists would waste all their time trying to solve it. The Monty Hall problem is the modern equivalent; it's computationally simple, but drop it into any group of people and they'll collapse into argument.)
I have two theories for why this particular puzzle is so hard to explain. The first is that it presents as a statistics problem, but it’s really a game theory problem. So here’s my own attempt to explain it:
You don’t know anything about the doors, so from your perspective, there’s a 1/3 chance the car is behind each door. (I think everyone can agree on that.) Let’s say you choose door 1.
Now Monty shows you there’s a goat behind door 2, and this is where things go off the rails.
If Monty flips a coin to pick which door he opens, and it happens to be a goat, then all you’ve learned is that the car isn’t behind 2. Doors 1 and 3 were equally likely before, so now they’re still equally likely, and it’s 50/50 whether you switch or stay. This is the obvious “statistics answer.”1
But put yourself in Monty’s shoes. He knows what’s behind each door, and he’s a game show host who wants to build suspense. Is he going to show you the car, leaving you with two goats? Of course not! He’s going keep the car hidden, so everyone still thinks you might win.2
So there are three ways the game could go:
The car is behind 1. It doesn’t matter which door Monty shows you; that door is a goat, and so is the remaining door.
The car is behind 2. Monty won’t show you 2, because that would ruin the game, so he shows you 3 (a goat).
The car is behind 3. Monty won’t show you 3, because that would ruin the game, so he shows you 2 (a goat).
Remember, there was originally a 1/3 chance the car was behind each door. So there’s now a 1/3 chance the car is behind your door (1), and a 2/3 chance the car is behind the other door that Monty didn’t show you.
So you’re better off switching, but not just because of statistics—it’s because you can predict how Monty will respond in each scenario.
Which leads to the second theory: switching feels wrong, so it’s hard to admit it’s better.3 There are well-known psychological biases toward keeping what we have instead of switching. But I’d go further and say those biases are even good.
Consider another scenario: You’re walking through a street market and see a game of three-card monte: one card is an ace, for which you’ll win $10; the other two cards are jokers, for which you’ll lose $5. You lose track of which card is which, so you choose one randomly. Then the host says, here, I’ll show you one of the cards you didn’t pick is a joker. Do you want to switch to the other card?
From a logic perspective, this is exactly the same as the Monty Hall problem. And yet the answer is the opposite: never ever switch!
That’s because you can also predict how the three-card monte host will act in each scenario, and it’s different: whereas host Monty didn’t have a stake in whether you won or lost and just wanted to make exciting television, host-monte only wants you to lose. So if you picked a losing card, he’s not going to offer to switch; he’ll take the money and run. If he’s offering you a chance to switch, it’s because you’re holding the ace.
Humans have spent a lot more time making zero-sum transactions than appearing on game shows, so it’s no surprise that our instincts are better suited to the former situation—and that the Monty Hall problem continues to flummox people.
![A cartoon image of the Monty Hall problem. [DALL-E]](https://substackcdn.com/image/fetch/f_auto,q_auto:good,fl_progressive:steep/https%3A%2F%2Fsubstack-post-media.s3.amazonaws.com%2Fpublic%2Fimages%2F8f621a4a-6507-43dc-ab49-92ea953b135d_1024x1024.webp "A cartoon image of the Monty Hall problem. [DALL-E]")
More formally:
From before, there’s a 1/3 chance the car is behind door 1. When that’s the case, Monty will show you a goat regardless of which door his coin chooses.
There’s a 2/3 chance the car is behind 2 or 3. When that’s the case, there’s a 50% chance Monty’s coin chooses the car door, and a 50% chance it chooses the goat door (in which case the car is behind the other door). So there’s a 50% * 2/3 = 1/3 chance that the car is behind 2 or 3 and he shows you the goat, and a 1/3 chance that he shows you the car.
Once you see that Monty shows you a goat, you can rule out the scenarios where he shows you the car. You’re left with the chance that your door has the car (1/3), or that one of the other doors has the car and he shows you a goat (1/3). Those are equal, so the new chance your door has the car is 50-50.
According to Hall, he didn’t always follow this logic, but he usually did.
If you want evidence: in one study only 13% of participants chose to switch. Even if you don’t understand the solution, you should think it’s a 50-50 bet, so there’s no purely rational reason to keep your original door so often.