16 April, 2009

The Monty Hall Problem

This is a strange one. I can't think of another theoretical maths problem quite like it. The answer is so counter-intuitive that generations of gifted mathematicians have got it wrong and undoubtedly generations to come will too. It goes like this:

Suppose you're on a game show and you're given the choice of three doors. Behind one door is a car; behind the others, goats. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 3 and the host opens Door 1, which has a goat. He then asks you "Do you want to switch to Door Number 2?" Is it to your advantage to change your choice?

So is it better to stick with your original choice or switch to the other unopened door? At first it seems the obvious answer is that it makes no difference what you do. Its a 50/50 choice, right? Well, no, wrong actually. You are always better off switching to the other unopened door. You might not believe it but it's been tested time and again and the conclusion is always the same.


I know what you're thinking. I thought the exact same thing. But the fact remains that I was wrong, and so are you. Here's why:

When you first make your choice it's a 1/3 chance that you have picked the door with the car behind it. When Monty opens one of the other doors and offers you the switch, by sticking with your original choice nothing has changed, so your odds are still 1/3. However, if you switch you are effectively getting both the opened door and the other unopened door, meaning yours odds have become 2/3.

Again, I know what you're thinking. "If I stick it's also a 2/3 chance because I now know what's behind one of the doors". But again, you would be wrong. Another way to think of it is to imagine 1,000,000 doors instead of just three. When you choose one door you have a 1/1,000,000 chance of winning. Monty then proceeds to open 999,998 of the other doors, all of which show goats, leaving just one other unopened door. If you now stick, your odds are still 1/1,000,000 because the new information makes no difference to the choice you originally had. If you switch your odds jump to 999,999/1,000,000, because you essentially have been given every door except your original choice. The same rationale holds for 3 doors.

The below figure should explain it:

Hopefully it now makes sense. If not, don't worry. You are in good company. As cognitive psychologist Massimo Piattelli-Palmarini says "... no other statistical puzzle comes so close to fooling all the people all the time" and "that even Nobel physicists systematically give the wrong answer, and that they insist on it, and they are ready to berate in print those who propose the right answer."


Anna Sethe said...

I'm not sure if you got that right. Sure, the conclusion is right, but I would calculate it differently:

There are two round of choices:
1. The probability that you're wrong is 2/3 while it's 1/3 for being right.
2. One door was opened, now you choose between two doors. The probability to be right is now 1/2.
But then you have two multiply the two steps of a probability tree. The first step is 2/3 vs. 1/3 the second step 1/2 vs 1/2, thus 2/3 x 1/2 vs 1/3 x 1/2
(Overall possibilities: 1st right(1/3) 2nd right(1/2); 1st right(1/3) 2nd wrong (1/2); 1st wrong (2/3) 2nd right(1/2); 1st wrong(2/3) 2nd wrong (1/2) adds up to one)

Kind of difficult to explain without a picture

Anonymous said...

This problem is also explained by an autistic boy in the book 'The Curious Incident of the Dog in the Night-time'.

rhiggs said...

Hi Anna,

I'm no expert but from what I've read, your method is the 'conditional' solution, and is of course mathematically correct. But my way still holds. According to Wiki, it is the 'popular' solution. I see it as a way of verbalising the 'conditional' method, but actually using the same underlying maths. But in fact, thats not quite correct, as it all depends on the set-up of the game.

According to Morgan et al. (1991) "The distinction between the conditional and unconditional situations here seems to confound many." That is, they, and some others, interpret the usual wording of the problem statement as asking about the conditional probability of winning given which door is opened by the host, as opposed to the overall or unconditional probability. These are mathematically different questions and can have different answers depending on how the host chooses which door to open when the player's initial choice is the car (Morgan et al., 1991; Gillman 1992). For example, if the host opens Door 3 whenever possible then the probability of winning by switching for players initially choosing Door 1 is 2/3 overall, but only 1/2 if the host opens Door 3. In its usual form the problem statement does not specify this detail of the host's behavior, making the answer that switching wins the car with probability 2/3 mathematically unjustified. Many commonly presented solutions address the unconditional probability, ignoring which door the host opens; Morgan et al. call these "false solutions" (1991). Others, such as Behrends (2008), conclude that "One must consider the matter with care to see that both analyses are correct."Intersting stuff!


Anna Sethe said...

too much maths in school...
It's so much easier for me to think with probability trees :)

FrodoSaves said...


I like your writing style. It's energetic and infectious, and that well-placed "read more" link made me want to do exactly that. I think you could definitely take a crack at this science writing stuff!

As regards the math, it's been several years since I did any of it, but it seems to me that the probability tree isn't correct because the game is apathetic as to whether your first choice was right, whereas the tree assumes that you chose correctly. I'm not sure if that makes sense.

PersonalFailure said...

I saw this in that blackjack movie, um, 21. Now I think I kinda almost understand it.

Which is really saying something about your writing, rhiggs, as I can just barely add and subtract.

rhiggs said...

Frodo and PF,

Thanks for the comments. Much appreciated! :)


Anonymous said...

I'd have a 100% chance of picking the car... because I wouldn't pick the one that was making noise... unless it's a dead goat behind the doors... that would suck even worse than a live goat.

Anonymous said...

What a surprise, then, to discover afterward that the car has a horn that sounds like a goat, while the live goats became silent in terror of it!

Stephen Wells said...

The best way to deal with the whole problem is just to consider- what are the odds that you guessed right first time? Makes it much clearer.

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