Saturday, April 24, 2004

Brainteaser: While searching for mind-stretching exercises, I came across this topic. It's a two year old post and probably people who were watching blogs two years ago have hashed this one out already, but it's new to me.

The scenario is this: You're a contestant on a Monte Hall-style game show, and you're shown three doors, behind one of which is Fabulous Prizes. You pick one. The host then opens up one of the two doors you didn't pick, and reveals the booby prize. So now there's two doors left.

The question is this: If he offers you the chance to switch doors, should you take it?

The surprising answer: Yes.

This is counterintuitive. Logic says that, from a probability perspective, it shouldn't make a difference. Two doors equals a fifty-fifty chance, right? So the door you're on is as good a door as any. Change, don't change, it makes no difference.

But that's forgetting one thing--when you selected the door, your chances of being right were only one in three. But if you switch doors, your chances of being right become one in two. You're increasing your chances of being right by almost 17%.

I'm sure you're still dubious. "The moment you open up one door, the odds of your door being right increase from 1 in 3 to being 1 in 2 automatically. It still doesn't make a difference which door you pick."

But in order for that to be true, he would have to open a random door. Remember--if there's three doors, and only one has a prize, there will always be at least one door without a prize, and that's the door he's going to open. So his opening the leftover door first does not actually change the odds of your initial selection. The information he's revealed does not inherently increase the likelihood of your choice being right--the fact that one of the other two doors was prizeless was a given when we made our initial selection. So our door is no more likely to be right now than it was before.

However, the door we DIDN'T choose has better odds. We know that, when showing a booby prize door, this door was not selected.

In other words, that door has a fifty percent chance of being the correct door, while my own door still only has a one in three chance.

Apparently testing bears this out.

UPDATE: I was thinking about how to reconcile the odds that the one door had 1 in 2 odds and the other door had 1 in 3 odds to come up with just one set of odds, when I realized I didn't need to.

For the same reason the odds are still 1 in 3 for the door you picked, the other door still has odds of 2 out of 3.

Think about it this way--imagine there were 10 doors, and you had opened 8 non-prize doors.

Odds are still only 1 in 10 that you're right, because nine out of ten times, it's going to be the other door that's right--only in ten times are you going to get the door right on the first try. So odds are 9 in 10 it's the other door.

Same way here. After opening one door, the other unselected door still embodies the 2 in 3 odds you were wrong.

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