Imagine yourself on Let's Make a Deal. Monty Hall has you onstage, and shows you the proverbial three doors. Behind one of them is a new car, and behind the other two are goats. He asks you to pick a door to open. You do so. Monty Hall then opens one of the doors you didn't pick to show you that that door concealed a goat. Of course he can always show you a goat, regardless of what's behind your door. (Assume for now that when Monty has a choice of doors to open, he picks one at random.) There are now only two doors left, one of them being the one you picked. Monty Hall then asks you if you would like to stick with your original choice, or switch to the remaining door.

The question is this: Does it matter if you switch or not? And if it does, should you switch?

(If by chance you don't already know this puzzle, then I strongly encourage you to take a moment and figure out the answer for yourself.)

The intuitive response is: No, of course it doesn't matter if you switch. There are two doors, so the odds must be an even fifty-fifty. The fact that there used to be three doors doesn't matter at this point. It's like flipping a coin and having it come up heads nine times in a row: What are the odds that the next flip will come up heads? The answer is one in two, because the past events don't influence the present state.

However, this situation happens to be a little different. The correct answer is that it does make a difference, and that you should adopt a policy of switching. If you switch doors, the odds of getting the new car are two in three.

If upon reading the above paragraph, your reaction is to strongly doubt its veracity (and perhaps even my intelligence), then let me tell you right now you are in excellent company. You're joined not just by the majority of people with average levels of math education — professional mathematicians are just as likely to have that reaction. Quite a number of them have embarrassed themselves publicly by insisting that that fifty-fifty answer has to be correct.

What follows is a collection of seven different ways to explain why this counter-intuitive fact is true. These explanations effectively all say the same thing, but from different points of view. Most of them will probably sound unconvincing at first. Our intuition is so sure of itself, it can be hard to get it to budge. In my experience, most people guess the wrong answer, and after doing so, they will continue to remain unmoved by explanations for a while. Eventually they reach a tipping point, and can start to get a handle on where their intuition has gone astray. My guess is that explanations don't help until after that point has been reached. Which may be why it's always a different explanation that finally convinces. Whatever the reason, I recommend that you try on all of the following explanations, in any order you like, until you find one that feels like it works for you.

(Then, for extra credit, once you feel you've completely aceepted the explanation, go back and reread the ones that didn't convince you, and try to figure out why.)

At the very beginning, when you make your first selection, you have a
one in three chance of getting the car. We think that when Monty Hall
opens another door, the odds should go up to one in two, but remember
that Monty can *always* open a door. So imagine that you're
still trying to make your first choice. You think to yourself: "Door
#1 has a one in three chance of being correct. However, after I
announce that choice, Monty Hall will expose a goat behind one of the
other doors, and at that point the odds of Door #1 being correct will
shoot up to one in two. However, it now occurs to me that I don't have
to wait for him to do this, because I already know for a fact that he
will do this. It is already certain knowledge. So the odds of Door #1
being right are already one in two." But you can apply the same
reasoning to Door #2 and Door #3, and get the same result. This will
leave you with a total of 150% for all three possibilities, and this
is obviously wrong. Therefore our assumption that Monty Hall showing
us a goat will change the odds must be wrong. Which means that the
door we select must still have a one in three chance of being the
winning door. Since there is only one other door remaining, the
probability of it hiding the car must be two in three.

At the beginning, when you pick a door for the first time, you have a one-in-three chance of picking the car, and a two-in-three chance of picking a goat. What happens next? Assume for the moment that you've picked the car. Monty Hall therefore can choose either of the two remaining doors to open, since they both conceal goats, and when you switch, you are guaranteed to wind up with the other goat. Now instead assume that you've picked one of the goats to begin with. Monty doesn't have a choice of doors now; he must expose the remaining goat. So when you switch, you are guaranteed to wind up with the car. It's the only thing left. Since you have a two-thirds chance of picking a goat at the beginning, always switching gives you the car two-thirds of the time at the end.

Imagine that instead of three, there are a hundred doors — one
car and 99 goats. Let's say that you pick Door #34. Monty Hall then
opens a different door to expose a goat, and then offers you the
choice to keep your choice, or switch to one of the remaining 98
doors. Let's assume that you choose to stick with Door #34. He then
exposes another goat, and then another, and another, and so on, each
time offering you the chance to switch, and every time you turn him
down. Finally, he opens a 98th door, showing you a 98th goat. The only
doors still closed are Door #34, the door you picked, and Door #78.
For the last time, Monty gives you the option of switching. If you
take a moment to think about it, you'll realize that you'd be a fool
not to switch at this point. Odds are 99% that you started by picking
a goat, and thus you have forced Monty to expose all the remaining
goats in order to keep the car concealed. The odds that the car
is *not* behind Door #78 is therefore 1%. Right? So at that
point you switch. But now imagine the same scenario with only ten
doors. After Monty Hall opens eight door and exposed eight goats, then
principle is less pointed but otherwise the same. The odds are nine in
ten in your favor if you switch. Now consider the case with only five
doors ... four doors ... then three doors. The principle continues to
hold: your odds are always increased by switching.

Imagine instead that first you make your pick of Door #1, and then,
without opening any doors, Monty Hall asks you if you wish to stick
with your original choice, or switch to Door #2 *and* Door #3.
That is, if you switch, you get the car if it's behind either of the
other two doors. It's pretty obvious in this case that switching
doubles your chances. But this is actually identical to the original
problem. The opening of one of the doors you didn't pick can be seen
as effectively "giving" you that door.

The key point to realize is that when you switch, you will always switch to the opposite of what you originally picked. You can't switch from a goat to a goat, because Monty Hall always removes one of the two goats from consideration before you switch. (And of course you can't switch from a car to a car, because there's only one car all along.) No matter what happens beforehand, at the time of your switch there is one goat and one car. So, if you start with the car, switching moves you to the goat, and if you start with a goat, you will switch to the car. Switching therefore reverses the odds of the original game, putting them in your favor. Because you have a two-in-three chance of starting with a goat, you have a two-in-three chance of ending with the car.

Our confusion comes from how we think the probability breaks down. We tend to imagine the situtation as being something like this:

Assume for the sake of concreteness that you start by picking Door #1, and Monty Hall chooses Door #3 to open and expose a goat. The possible outcomes from that point are:

#1 #2 #3 car goat goat — you will win if you don't switch goat car goat — you will win if you do switch goat goat car — impossible, since Monty will never expose the car

This line of reasoning makes it look like there are only two possible situations, equally likely, and thus there is a fifty-fifty chance of winning. But this is faulty, because it fails to take into account that the door you start with affects the probabilities of what happens next. Your initial choice affects which door Monty Hall can choose. In fact, Monty doesn't even have a choice most of the time, and the probability table up above fails to take that imbalance into account. So instead, we must go back to the point at which you first selected Door #1, and break down the possible scenarios from there:

#1 | #2 | #3 | |

car | goat | goat | — Monty can open #2 or #3, and (either way) you will lose if you switch |

goat | car | goat | — Monty must open #3, and you will win if you switch |

goat | goat | car | — Monty must open #2, and you will win if you switch |

Thus we see that a policy of switching gives us the car two times out of three.

The door that Monty Hall chooses to open doesn't tell you anything
about the door you originally selected, but it does tell you something
about the door he *didn't* open — the door that he likely
*couldn't* open.

Texts

Brian Raiter