The Monty Hall Problem In Pictures
The Monty Hall problem is one of my favorite probability problems, because the solution seems so obvious, but turns out to be wrong. It is a hard problem to understand intuitively. Even once I saw the solution, I saw the math was correct, but didn't feel it intuitively.
This post will explain the problem, along with a solution that makes intuitive sense to me. I hope to tell a story that will help you believe the counter-intuitive solution to this problem.
Here's how it goes. You have three doors:
Behind two of the doors are sheep:
Behind one door is a car:
Your goal is to win the car.
I ask you to pick a door. Let's say you pick door one:
There's a 1 in 3 chance that that door has the car.
I know what's behind the doors of course, and now I open one of the other doors to reveal a sheep:
Then here comes the twist: I say "would you like to switch to the other door?"
So you can stick with door one or you can switch to door three:
What would you do?
You're trying to win the car. There's a 1/3 chance that the door you picked has the car behind it. The question is, when I revealed the sheep, did I change the odds? Is there any reason to think that door three is more likely to have the car than door one? I think most people would say no. That's what I did the first time I heard this problem, that it doesn't matter what you choose. There are two doors remaining, and they are both equally likely to contain the car -- that was my reasoning. And it was wrong!
Let me frame the problem in a different way. Let's play this game with 100 doors to start. Out of these 100 doors, one contains a car and the rest contains sheep:
You pick a door:
It's the same set up as before, except there's a lot more doors with sheep behind it. The odds that you picked the door with the car are even worse! 1 in 100. There's only a 1% chance you will win the car.
Then I offer you a great deal: you can stick with your door, or you can choose the entire set of the remaining 99 doors! So if the car is behind any of those 99 doors, you win the car:
What a great deal! Of course you switch. You've just gone from a 1% chance of success to a 99% chance of success. You start planning a road trip.
Here come the sheep
So you've decided to switch to the set of 99 doors. Now I open 98 of those 99 doors to reveal sheep:
There's one door left unopened. I offer you the same deal again. You can stick with your one initial door, or you can stick with the set of 99 doors. Exactly the same, except now you can see that 98 open doors that contain sheep. Has that information changed anything about the probabilities?
Remember, earlier we were in a similar situation, where we had only two closed doors. Then we said the odds are 50-50:
Would you say the same applies here?
Of course not. It's still 99 doors vs 1. You already knew most of those doors contained sheep -- at least 98 doors, in fact. Just because I showed the sheep to you, that's not giving you any new information.
Remember: I know where the sheep are! I know exactly which doors to open. I'm not opening doors at random. That's why opening doors isn't giving you any new information. That may not make a lot of sense to you, but this point happens to be key to the Monty Hall problem, as you will see in another game at the end of the post.
But back to the game. The set of 99 doors is still better. I mean, look at how many more doors that is:
You're happy with your choice, you're confident the road trip is still on. You start creating a road trip playlist in your head. You have a set of 99 doors! You can't lose.
Back to two doors
Oof, bad news, the producers of Monty Hall just called. I can't offer you the set of 99 doors after all. "We're not made of imaginary doors," is what they said. Here's an exact quote from a producer:
I’m like a door. Open to new things. EXCEPT LETTING CONTESTANTS PICK A WHOLE SET OF DOORS! phone slams
They said I have to offer you a choice between two doors, sorry. You can either stick with the door you initially chose, or you can switch to the one closed door in the set of 99 doors:
So you don't get the full set of 99 doors, you just get one other door. Now you are back to two doors, and the road trip dream is gone. Or is it?
Seems wrong, I know, but is it possible (99 closed doors - 98 open doors with sheep) == 1 closed door?
Back to Monty Hall
Our door manufacturer is running out of lumber, so let's go back to the original Monty Hall with just three doors. We can play this game:
You know what, I am feeling bad about your road trip, and the producers are on vacation, so I'll offer you a set again:
Isn't that generous of me? Here, I'll offer you the set a different way if you want:
But wait ... is there a difference between these games? For each of these games, would you stick with the door one or switch to door three?
Earlier it had seemed like the choice was 50-50:
But really, you can think of this as one door versus a set of two:
In that case, the probabilities are 1/3 and 2/3:
This turns out to be the correct answer. Yes, the probability of that set is really 2/3, even though one of the doors has a sheep! You should always switch, because there's a 1/3 chance your original door had the car, but a 2/3 chance the other door had the car!
I hope this post has fried your brain a little bit, and made you feel like there is more to this problem than meets the eye. Probability has a funny way of going against our intuition a lot of the time.
There are many related ideas I haven't covered here. I'll leave a couple for you to noodle over. Warning: they may keep you up at night:
- Suppose right after I reveal a sheep, a friend of yours walks into the room. They don't know what door you initially picked. What are the probabilities of the two doors for them?
- Suppose 1000 tickets have been issued for a lottery. You buy one ticket. You have a one in 1000 chance of winning. As they start checking tickets, none of the tickets is the winning ticket. Eventually the organizers check every ticket except yours and one other person's ... lets call them Aman. Your ticket has a 1/1000 chance of success, and the rest of the tickets combined have a 999/1000 chance of success -- that is, it's almost guaranteed someone else has the winning ticket. But now they have checked 998 tickets, and none of them were the winning ticket, which is just like opening doors to reveal sheep. In that set, Aman's ticket is the only one left. According to the solution above, Aman's ticket therefore has a 999/1000 chance of success. You should switch with him if he'll let you! Except from his perspective, the situation is exactly the opposite. The ticket he picked had a 1/1000 chance of success, and he thinks your ticket has a 999/1000 chance of success. Sometimes it really seems like the grass is greener on the other side. Who is correct?
Here's a hint: remember when I said earlier I know which doors to open to reveal sheep, and that is really important? This is different, because the lottery organizers don't know who has the winning ticket. In this case, each additional ticket they check is giving you new information.
Monty Hall tends to generate a lot of anger. If you think my explanation is wrong and are feeling angry at me, instead of emailing me, may I suggest making a donation to a local charity. Then you can rest knowing you have been a better person than me that day. Or donate to one of these fine organizations:
I liked this quote from the Wikipedia article:
Pigeons repeatedly exposed to the problem show that they rapidly learn to always switch, unlike humans.