This is your brain on teasers

Here's an oldie, but goodie.

Suppose you're on "Let's Make A Deal" and you're confronted with the notorious three doors. You get to pick one.

You make your choice, and before your chosen door opens, Monty Hall opens one of the doors you didn't pick to reveal a booby prize. He then asks you if you want to switch to the remaining door you didn't pick.

The question is, should you switch, should you stay put, or does it just not matter?

Dean has his analysis on the topic, as does Patterico, as does xrlq. (Hat tip chain, there.)

One of the comments to Patterico's post provides an analysis very similar in feel to a probability tree analysis.

Probability is a matter of counting. Calculating the probability is nothing more than counting up the number of alternatives to whatever event we're studying.

For example, if you roll a six-sided die, there is one way of getting a "4", and six possible ways the die can come up. The chance of getting a "4" is one in six. The chance that a prime number will come up will be either 1/2 or 2/3, depending on whether we count 1 as a prime.

In the Monty Hall problem, we need to count the number of ways of making our selection, and how many of them lead to a win. It's all in counting the right things. Rather than counting doors, let's count pathways to a win or a loss.

There are two choices the player makes.

The first choice is door number 1, 2, or 3.

After the player makes his choice, Monty opens one of the remaining doors, behind which is one of the booby prizes.

Now the remaining choice is, switch to the remaining door, or stay put.

The trick here is, Monty knows where the grand prize is hiding.

The player has a 1/3 chance of picking that door on his first try. If he picks that door, it doesn't matter which door Monty picks. If the player switches, he loses; if he stays put, he wins.

In the other two cases, the player picks a losing door. Monty is not going to open the door with the grand prize behind it, so he will open the other booby prize door with a probability of 100%. Monty's choice has no impact on the odds, because he really has no choice.

The player is sitting on a losing door, and if he switches, the only option left is the door with the grand prize behind it.

This will happen two thirds of the time.

If the player makes the first selection at random, and always switches when offered the chance, he will win the grand prize two times out of three.

And this ignores what I think is very likely to be true – the first choice of the three doors is not random. I'm sure one number is more likely to be chosen than the others. But even in the case of a random choice, letting Monty open one door, and then switching to the other one turns a 1/3 chance of winning into a 2/3 chance.