"Let's Make A Deal" Explained

There is a woman named Marilyn vos Savant (must be a pseudonym, since ‘savant’ means ‘very smart’) who claims to be the smartest person in the world, and has no problem letting you know that. She writes a widely syndicated column where readers can ask her anything they want, and she will answer.

A few years ago, somebody sent her the question about the Let’s Make a Deal scenario. Ms. Savant replied that the odds of winning the prize if you change your mind are twice those if you don’t.

For this, Ms. Savant was roundly ridiculed. Physicists, professional mathematicians and all sorts of other very smart people said in print that she was nuts.

They all ate their words. She was right.

I’m going to explain this a bunch of different ways, in the hope that one of them hits you with the "Aha!" effect.

So here goes.

1- Suppose I revised the game slightly, as follows.

There are three doors. A, B and C

There is a dividing line on the stage.

You can put any two doors on one side of the line, and the remaining two on the other side, like this:

A B | C

(Since the location of the prize is random, it really doesn’t make a difference which doors you put where.)

Now, here’s your choice: you can pick the two doors, or you can pick the one door.

If you pick the lone door, you win if the prize is behind that door.

But if you pick the two doors and the prize is behind either one, you win.

What’s your choice?

Blindingly obvious, you say. I’d pick the two doors, of course.

Of course. Your odds of winning are twice that of picking only one door.

This situation is exactly the same as the game I described.

When you pick one door (let’s say A), the other two (B and C) go on the other side of the line. By opening the empty one of those two doors, and letting you change your mind to the other still-closed door, that’s the same as letting you pick both doors together. Except instead of opening both doors at the same time, Monty opened one of them in advance.

In other words, you may think you’re picking one door and then changing your mind to one other door, but what’s really happening is that you’re picking one door to start, and then being allowed to pick both the other doors. You’re going to win the prize if either of those two doors contains it. Seeing which of the two doesn’t contain it first is just a bit of meaningless misdirection.

2- Didn’t like that? No problem. Try this...

There are only two variables to this game:

a) Which door contains the prize?

b) Do you change your mind or not?

So let’s list all the possible outcomes. (I hope you already agree that it doesn’t matter which doors go where, since the prize is placed at random.)

Here is a little grid of all the possibilities:

Prize is behind:

There are only six different ways things can turn out.

Now, let’s assume that we always pick A as our first door. (If this bothers you, let’s just pretend that we don’t label the doors A, B and C until you’ve made your choice. All the possibilities are really the same.) We can now fill in the grid with the outcomes, Win or Lose, for each possible scenario.

For example, if the prize is behind Door A (which you always pick first) and you don’t change your mind, that’s a Win. If you did change your mind, that’s a Lose.

So here’s what the grid looks like when it’s all filled in:

Prize is behind:

It is easy to see that, if you change your mind every time, there are two ways to win, and only one way to lose. If the prize is behind A, you lose. But you win if it is behind B or C.

However, if you don’t change your mind, there is only one way you can win, and that is if the prize is behind A. If it’s behind B or C, you lose.

So changing your mind gives you twice as many ways to win.

Remember: think of A as being the door you chose first, regardless of which one you really chose, and B and C as the other two. That way, you won’t have to draw a separate grid depending on whether you chose B or C first instead of A.

But if it makes you feel better, go ahead. The results will be the same.

3- Didn’t like that either, eh? Okay.

Instead of three doors, let’s pretend that there are a thousand doors.

You pick one, and the other 999 go on the other side of the line.

A | B C D E F G H I J K L M N O P ..... (etc.)

Now Monty opens up all the empty ones on the right side of the line except one:

A | X

Now Monty tells you you can change your mind.

Pretty obvious that the odds of the prize being behind X are 999-to-one.

Or, putting it another way, the odds that the very first door you picked, A, contains the prize is 999-to-one against you.

It’s the same situation as in the original game. When you selected A to start with, you only had a one-out-of-three chance of picking the right door. There was a two-out-of-three chance that it was in either B or C. Just because Monty told you which of B or C the prize wasn’t in doesn’t change the odds of the prize being on that side of the line, just as if there had been a thousand boses to start with.

4- One last go, a little differently. This one gets to the heart of the matter, which is our reflex insistence that, if there are two choices, they both have equal probabilities.

You may be having trouble getting over the fact that, no matter what else has gone on, when you’re left with only two doors, there’s a 50/50 chance the prize is behind one or the other, right?

A | B

Stop and think for a second: why do you assume that the probability is 50% each?

Well, there are two doors, and they each have an equal probability of containing the prize. What could be simpler?

Suppose I were to tell you that the television show’s vice-president-in-charge-of-prize-hiding really likes to put the prize behind B most of the time. In fact, he does it 9 times out of 10.

Now what are the probabilities?

A is 10%, and B is 90%.

So there’s one way it could be other than 50/50. And there are plenty of others.

So forget the notion that two doors necessarily means 50/50 odds. That’s only true if you only had two doors to start and the prize placement was random.

Now I’ll tell you that, before we got down to Door A and Door B, there used to be two doors on the A side. And each of those doors had a 1/3 chance of containing the prize.

Now something is different. The odds that the prize was in either A or the other door (C) are twice the odds of it having been in B. So when C disappears, and we know for a fact that the prize wasn’t there, the probability that it is in the remaining door A is still 2/3.

And if it makes you feel better, leave door C sitting right there, open, instead of taking it away. Now do you get it?

If not, I apologize: I’m out of explanations.

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