The Monty Hall Problem is one that comes up fairly regularly in alt.folklore.urban, and it is one that people consistently have problems understanding, hence it is only reasonable to have a sort of FAQ for it. This FAQ is organized thusly: Part 1: The Monty Hall Problem itself, and the Answer Part 2: Compelling Explanations of the Answer Part 3: Common Misconstructions of the Problem, and their Rebuttals

You are a guest on Let's Make a Deal, hosted by Monty Hall. He presents you with three doors, behind one of which is a Brand! New! Car! and behind the other two are goats. You are given one guess as to which door has the car. Monty then opens one of the other doors to show you a goat, and you are given the opportunity to switch to the other unopened door. Should you switch, or should you stay with your initial guess?

Note: There are certain assumptions in the problem that are important to understand. One is that Monty is honest, and always behaves the same way. This means that you are always provided the choice to switch, which is apparently not the case on the actual Let's Make a Deal show, and that Monty's subsequent behavior does not change relative to whether you initially picked the winning door. This is a problem about probability analysis, not about realistically modeling the behavior of a game show host.

And the answer is... yes, you should switch. Your odds of winning increase to 2:1 if you switch, and remain at 1:2 if you do not.

This seems counterintuitive to most people. The seemingly obvious logic goes something like, "I now have two doors, each of which could have the car. The odds that each door has the car is 1:1, or 50%, so it makes no difference if I stay or switch." It is compelling logic on the surface, but it ignores an incredibly important part of the problem: Monty knows where the car is.

Why is this important? Because when Monty opens a door before offering you the chance to switch, he never opens the door with the car. This changes what is called the "problem space" and keeps it from being merely an exercise about random guesses.

Part 2: Compelling Explanations of the Answer

I have found that in explaining Monty Hall, different people respond to different ways of restating it. One explanation works for Ernest, who now understands, but it doesn't make things any clearer for Miguel, who responds to a different way of looking at the problem. The idea here is to show a few different ways to understand it, so that any one of them might help one understand why the odds shift the way they do.

1. The essence of the choice can be boiled down to this: you get one door, which leaves Monty with two doors. You know that at least one of Monty's doors is a loser, so him showing you that fact changes nothing. Monty offers you the choice of either having your one door, or his two doors (a loser of which he already opened for you). His two have a collective 66.6% chance of winning and your one has a 33.3% chance of winning.

2. Suppose there were a thousand doors, and you chose one. Monty then keeps opening door after door, and you stay with your original door each time. He opens 998 doors, all of them goats. Do you really think the odds are 50% you chose the correct one door out of the thousand right off the bat? You'd very seriously consider switching in this situation, wouldn't you? Doesn’t it seem obvious that Monty's 999 doors have a better chance of having the winning door in there someplace?

This is meant to show that it's easier to see the odds shift when the numbers are larger. The odds in this example are obviously much better than in the real problem, but the mechanic is the same: switching increases your odds of winning, and this is an exaggerated example of that.

3. There are basically three ways this can go down. Assume Door 1 is the winner:

A. You pick Door 1. Monty shows you another door and you switch. You lose.

B. You pick Door 2, which is a goat. Monty shows you Door 3 and you switch. You win.

C. You pick Door 3, which is a goat. Monty shows you Door 2 and you switch. You win.

Those are the only real ways it can play out. Two times out of three (B and C), switching will win you the car.

4. Let's just brute force the thing for every explicit combination, and tally up the odds.

Part One: There are these doors, see, and you pick one. There are nine scenarios:

You pick Door 1, Door 1 has the Car (12.5% chance)

You pick Door 1, Door 2 has the Car (12.5% chance)

You pick Door 1, Door 3 has the Car (12.5% chance)

You pick Door 2, Door 1 has the Car (12.5% chance)

You pick Door 2, Door 2 has the Car (12.5% chance)

You pick Door 2, Door 3 has the Car (12.5% chance)

You pick Door 3, Door 1 has the Car (12.5% chance)

You pick Door 3, Door 2 has the Car (12.5% chance)

You pick Door 3, Door 3 has the Car (12.5% chance)

Part Two: Monty shows you a door and offers a switch. There are only 12 possible scenarios (more on that below):

You pick Door 1, Door 1 has the Car, Monty opens Door 2 (6.25% chance)

You pick Door 1, Door 1 has the Car, Monty opens Door 3 (6.25% chance)

You pick Door 1, Door 2 has the Car, Monty opens Door 3 (12.5% chance)

You pick Door 1, Door 3 has the Car, Monty opens Door 2 (12.5% chance)

You pick Door 2, Door 1 has the Car, Monty opens Door 3 (12.5% chance)

You pick Door 2, Door 2 has the Car, Monty opens Door 1 (6.25% chance)

You pick Door 2, Door 2 has the Car, Monty opens Door 3 (6.25% chance)

You pick Door 2, Door 3 has the Car, Monty opens Door 1 (12.5% chance)

You pick Door 3, Door 1 has the Car, Monty opens Door 2 (12.5% chance)

You pick Door 3, Door 2 has the Car, Monty opens Door 1 (12.5% chance)

You pick Door 3, Door 3 has the Car, Monty opens Door 1 (6.25% chance)

You pick Door 3, Door 3 has the Car, Monty opens Door 2 (6.25% chance)

There are now two closed doors, and you're asked if you want to switch. Total up the currently winning scenarios above, and they amount to 6:12, or 1:2. The scenarios involving the door you didn't initially pick total up to 6:3, or 2:1.

Note: If you're confused as to why some choices have a 6.25% chance and others have a 12.5% chance, it's simple. There are 18 mathematical scenarios, but Monty won't let six of them happen. You'll never see these six play out:

You pick Door 1, Door 2 has the Car, Monty opens Door 2

You pick Door 1, Door 3 has the Car, Monty opens Door 3

You pick Door 2, Door 1 has the Car, Monty opens Door 1

You pick Door 2, Door 3 has the Car, Monty opens Door 3

You pick Door 3, Door 1 has the Car, Monty opens Door 1

You pick Door 3, Door 2 has the Car, Monty opens Door 2

Remember, he only opens a loser, it makes no sense to open the winner, and Monty knows where the winner is.

5. If you like mathematical proofs, this problem can be proved by Bayes' Theorem. An excellent explanation of Bayes' Theorem can be found at the Finite Mathematics & Applied Calculus Resource Page's chapter on Bayes' Theorem, and also at the site is a nice Java app that demonstrates the Monty Hall problem.

Q, as they say, ED. Like I said, different people will realize the truth of the answer different ways, so having just one way to describe the odds is not the best strategy for explaining it. Methods 1 and 2 above are, I've found, the best for getting the epiphany to strike in your average person, but it can be hard to tell what people will respond to.

Part 3: Common Misconstructions of the Problem, and their Rebuttals

Q. You randomly picked a door at the beginning, and now you're randomly deciding that the other door is better. It's not an informed choice, it's just like flipping a coin, so it's fifty-fifty.

A. You randomly picked, and your odds were 1:2 that your random guess was right. Just because Monty showed you something you already knew, that one of his doors is definitely a loser, doesn't increase the odds that your initial guess was right.

Q. Monty doesn't allow you to switch to either of the two unpicked doors, he only allows you to switch to one of the remaining doors. Because he constrains your choice, it's a simple matter of one prize and two doors, fifty-fifty.

A. That's not actually the nature of the choice. Given that we know the unpicked two doors have a total probability of 2:1 of containing the car, let's examine how that breaks down. We know that one door has to be a loser, right? He can't have two winners. So we know that, instead of each door having a 1:2 chance of winning, one door has a 0:3 chance of winning and the other has a 2:1 chance of winning (the aggregate odds don't split equally between the doors). Monty is always going to show you a 0:3 door. He never actually changes your odds by constraining your choice of doors to switch to.

Q. All doors have a 1:2 chance of winning at all times. You can never have more than a 1:2 chance of winning.

A. This is true if you only just pick a door and the game is over. By this logic, one should agree that if one door has a 1:2 chance of winning, then two doors have a 2:1 chance of winning. You have one door and Monty has two, and his choice is, in essence, offering you his two doors for your one. Oh, and he shows you that one of his doors is a loser, but you knew that already. He's still offering to trade his two doors for your one door, or trade his 2:1 chance for your 1:2 chance.

Q. Once Monty opens a door, it's gone, and may as well never have been there. It's now just two doors, and the odds are fifty-fifty.

A. No, because when you choose to switch to the other door, you're not selecting a single door, you're selecting the entire set of doors you did not initially pick, and the odds that go along with them. You just happen to know now which doors of that set are losers. If you picked one of ten doors, and he offered that you could keep your door or take his nine doors instead, what would you do?

Q. All this odds business works out only over a lot of tries. In the real world, it's one try, two doors, and that means 50%.

A. Odds do not work this way. The odds do not change because it is one try or one million. The only thing you can say about it being only one try is that you can't really draw an accurate conclusion about what the pattern would end up being going solely by one empirical attempt. The odds for each isolated instance of the problem will always be the same: 33.3% if you stay, 66.6% if you switch, and while this becomes more visible the more times you take the test, it is still the case for every single instance as well.

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