Let's Make a Deal is famous in part for having within it a very curious game element. That element has come to be known as the Monty Hall Problem after the name of the host of the show.

The basic idea is that you're presented with three doors, Door A, Door B, and Door C. Behind one of those doors is something pretty cool. It could be some money, or a car, or whatever. Behind the other two doors there is something less cool, like a goat.

Unless you really like goats, in which case the goat is the prize and the other doors have something that you find less cool. Like less cool goats.

By the way, thanks as usual to wikipedia for having a pretty sweet totally public domain image to put things into perspective.

So the host brings you up to these doors, and tells you there is something great behind one of them. Without anything but dumb luck to guide you, he tells you that you're allowed to pick a door and receive the prize behind it.

Pretty straightforward, right? You have a 1 in 3 (1/3) chance of picking the good prize, and a 2 in 3 (2/3) chance of walking home with a goat (assuming from this point forward that you're one of the people who is trying to

*not*win a goat).

This should make sense to you - this is the easy part. Don't forget this part, though, as the rest turns out to be about that easy.

You pick a door. So, pick a door. Take a deep breath, and wonder if you've won. You look at the host, and he's smiling. You start to get the feeling he knows something that you don't.

Well, it turns out that he does. In fact - for all intents and purposes - he knows EVERYTHING. Sure, he can be easily blinded to the situation by receiving information in an earpiece, but it's so much more fun if he's really just trying to have some fun with his all-knowingness.

The host informs you that you have two options. You can either stick with the door you just picked, or switch to one of the others. You think about it for a moment and realize that either of the other two doors has the same odds of a prize as the one you just picked. No reason to switch or not switch, as you may as well have flipped a coin in the first place.

Wait - the host says - that's not all. He points at a door that you haven't picked, and it opens. Hey, there's a goat behind it! He has just revealed one of the losing choices. The only rule is that he has to pick a door that you didn't pick.

For example, if you picked Door A, he may open and reveal a goat behind Door B or Door C. If you picked B he may open and reveal a goat behind Door A or Door C. If you picked Door C he may open and reveal a goat behind Door A or Door B.

It's not his choice which door gets opened in all situations - no matter what is behind the door you picked there is still at least one goat remaining behind one of the other doors. If you picked the winner right away he has a choice of two goats, but if there is a goat behind your door then there's still a goat remaining on the board that he can reveal.

This is where it gets interesting.

The host asks if you want to stay with the door you picked first, or switch to the last remaining door. You ponder it for a moment. What's the benefit of switching? You already decided that all of the doors are a coin flip anyway, right? Right? Right...?

Let's walk through a possible scenario.

Let's say the correct door is Door C. The host knows it, but you don't, and you can't learn it. Here's how things play out if you stay:

You initially pick door A. The host reveals door B, but you stay with A. You lose.

You initially pick door B. The host reveals door A, but you stay with B. You lose.

You initially pick door C. The host reveals door A, but you stay with C. You win.

In the case of staying, you have to correctly guess the right door out of three

*on your initial try*. You have a 1 in 3 chance of winning, because you pass on the second step. You already figured this out as the easy part earlier in the post. Now what happens if you switch?

Here's how things play out if you switch:

You initially pick door A. The host reveals door B, and you switch to the door remaining, C. You win.

You initially pick door B. The host reveals door A, and you switch to the door remaining, C. You win.

You initially pick door C. The host reveals door A, and you switch to the door remaining, B. You lose.

You see, when switching, you're actually hoping that you

*chose poorly*on your first guess.

If you picked wrong to start, you'll have the correct door when you switch. Do you see that part? That's the trick.

When you get to the second choice - when you have an opportunity to switch - you're playing a slightly different game. There's something cool behind one door, and a goat behind the other. One of the doors (the one opened) is out of play. The door that's out of play was part of the pair involving the door that you can switch to - in essence you're not switching to one door, you're switching to both of the doors you didn't pick initially. It's just that one of those doors has already been opened.

Think of it this way. Instead of opening a door and then asking you if you want the remaining, the host is actually asking you if you want to stay with your initial door choice or switch to

*both of the other doors*.

One of those doors has a goat, and the host knows it. He actually likes goats, so he'll simply open that door and take the goat off your hands. He'll never steal the cool prize, because he's not like that. You get whatever is in the door that he didn't choose.

He may leave you a goat (if you picked right and have the cool prize behind your door), or leave you the cool prize (if you picked wrong and have a goat behind your door).

Since there are three doors you have a 2 out of 3 chance of picking incorrectly to start. If you stay, you need to have picked the right door out of three - and there you have your 1 in three chance.

On the surface people assume you always have a coin flip choice (or overweight their initial guess), while the best bet is to always switch. Over time, you'll win more prizes (and less goats).

I frequently talk about this problem at parties (well, sort-of-geeky parties), family reunions etc and even though I use pretty much the same reasoning as you did, there are often people who just won't beleive.

ReplyDeleteA nice way to convince them is using an extended (and more party-friendly) version of Monty Hall: take a pack of cards, deal them face down and ask a non-beleiver to guess which is the ace of spades. After they choose, reveal all the other cards but one and ask them whether they want to switch or not.

I think the key problem here is the inability of people to see the first choice and the second choice as independent. I teach this in my statistics class and even the students who "get it" still feel like it is incorrect on a gut level.

ReplyDeleteI used a similar trick to what anonymous did to show this to my parents and it worked rather well. In class, I make students play 20 rounds of the game with a partner. For ten of the rounds no matter what, they have to switch and for ten of the rounds no matter what, they have to stay. Even if they don't get it initially they seem to understand it after this game because to fully appreciate probability you have to have more than one trial! (of course).