the Monty Hall problem

The apparent winning strategy defies intuition. Jason Rosenhouse describes the game:

The Monty Hall problem goes like this: You are presented with three identical doors. Behind one of them is a car and behind the other two are goats. You want the car. Monty Hall tells you to choose one of the doors. Regardless of which door you choose, at least one of the two remaining doors will have a goat behind it. Monty Hall, who knows where the car is, then opens one of the doors that has a goat behind it. He then gives you the option of either sticking with the first door you chose, or switching your choice to the other unopened door.

Question: What should you do? Should you stay where you are? Switch? Does it make a difference? [emphasis added]

I first read about the classic probability conundrum in a Marilyn Vos Savant column. (Yes, that's her real name. Scroll down to read about the column. Some other time, visit the obsessive fact-checker, Marilyn is Wrong.)

People generally think that when Hall reveals a wrong answer, the odds then change to 1 out of 2; after all, there are two doors left, and te car has to be behind one of them. But they don't. In fact, 2 out of three times, the player should switch in order to win. Intuition says it shouldn't matter. Intuition is wrong.

The mathematician writes,

Try this computer simulation. Do it many times and keep track of your statistics.

Think the computer simulation is rigged? Fine. Get out a pad and a pencil. Make a list of every possible scenario. (For example: Car behind door number one, you choose door number one, Monty Hall opens door number two.) It's a little tedious, but there aren't that many possibilities. Then put a little mark next to all the scenarios in which you will win by switching. I think you will find that by switching you will win 2/3 of the time.

Since you're still skeptical but lazy, I'll take on Rosenhouse's challenge for you, listing every single possibility right here, right now.

It's in | You choose | Hall opens | You choose to... | OUTCOME
Door 1. Door 1. Door 2. Stay. WIN
Door 1. Door 1. Door 2. Switch. LOSE
Door 1. Door 1. Door 3. Stay. WIN
Door 1. Door 1. Door 3. Switch. LOSE
Door 1. Door 2. Door 3. Stay. LOSE
Door 1. Door 2. Door 3. Switch. WIN
Door 1. Door 3. Door 2. Stay. LOSE
Door 1. Door 3. Door 2. Switch. WIN
Door 2. Door 1. Door 3. Stay. LOSE
Door 2. Door 1. Door 3. Switch. WIN
Door 2. Door 2. Door 1. Stay. WIN
Door 2. Door 2. Door 1. Switch. LOSE
Door 2. Door 2. Door 3. Stay. WIN
Door 2. Door 2. Door 3. Switch. LOSE
Door 2. Door 3. Door 1. Stay. LOSE
Door 2. Door 3. Door 1. Switch. WIN
Door 3. Door 1. Door 2. Stay. LOSE
Door 3. Door 1. Door 2. Switch. WIN
Door 3. Door 2. Door 1. Stay. LOSE
Door 3. Door 2. Door 1. Switch. WIN
Door 3. Door 3. Door 1. Stay. WIN
Door 3. Door 3. Door 1. Switch. LOSE
Door 3. Door 3. Door 2. Stay. WIN
Door 3. Door 3. Door 2. Switch. LOSE

Count up your STAY wins and losses. 6-6
Count up your SWITCH wins and losses. 6-6.

Hence, the probability is 1/2, not 2/3, right?

Wrong.

What you might have forgotten when listing all the outcomes is that initially choosing the correct door is always a 1 out of 3 likelihood--so the probability that Hall will open Wrong1 or Wrong2 isn't 1/4, as the list would suggest, but 1/6 (1/2 * 1/3).

If you're not mathematically minded--and I'm not--it's easy to get confused.

Here's a better way to show the results for Door 1.

It's in | You choose | Hall opens | You choose to... | OUTCOME
Door 1. Door 1. Door 2 or 3. Stay. WIN
Door 1. Door 1. Door 2 or 3. Switch. LOSE
Door 1. Door 2. Door 3. Stay. LOSE
Door 1. Door 2. Door 3. Switch. WIN
Door 1. Door 3. Door 2. Stay. LOSE
Door 1. Door 3. Door 2. Switch. WIN

Now you can see that the numbers balance out properly, and that the odds are in your favor to switch.

Incidentally, a game that lets its contestants win 2/3 of the time is a pretty stupid game.