# The Monty Hall Problem

Here’s how Monty’s deal works, in the math problem, anyway. (On the real show it was a bit messier.) He shows you three closed doors, with a car behind one and a goat behind each of the others. If you open the one with the car, you win it. You start by picking a door, but before it’s opened Monty will always open another door to reveal a goat. Then he’ll let you open either remaining door.

Suppose you start by picking Door 1, and Monty opens Door 3 to reveal a goat. Now what should you do? Stick with Door 1 or switch to Door 2?

[...]

This answer goes against our intuition that, with two unopened doors left, the odds are 50-50 that the car is behind one of them. But when you stick with Door 1, you’ll win only if your original choice was correct, which happens only 1 in 3 times on average. If you switch, you’ll win whenever your original choice was wrong, which happens 2 out of 3 times.

## Notes:

You should always switch doors because you had a 1 in 3 chance of getting the right one the first time, and a 1 in 2 chance if you switch.

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**And Behind Door No. 1, a Fatal Flaw**

**Electronic/World Wide Web>**

**Internet Article:**Tierney, John (April 8, 2008)

*, And Behind Door No. 1, a Fatal Flaw*, New York Times, Retrieved on 2013-04-08

**Folksonomies:**games mathematics cognitive dissonance