jordanlund, 1 year ago

The Monty Hall problem.

You are given a choice of three doors, let’s call them 1, 2, and 3.

Behind one of the doors is a fabulous prize. Behind the other two are joke prizes worth nothing.

You are asked to pick a door. It doesn’t matter which one you choose, because it’s not opened inmediately.

Instead, the host opens one of the doors you did not pick to reveal the gag gift.

He then asks you if you want to change your choice.

What are the chances of winning? Should you choose a different door, or keep your existing choice?

The math says, your chance of winning if you stay with your choice is 1/3. Revealing the contents of one door does not change that, it’s still 1/3.

Switching to the other door gives you a 2/3 chance of winning. Not 1/2 or 1/3.

behavioralscientist.org/steven-pinker-rationality…

“If the car is behind Door 1, you lose. If the car is behind Door 2, Monty would have opened Door 3, so you would switch to Door 2 and win. If the car is behind Door 3, he would have opened Door 2, so you would switch to Door 3 and win. The odds of winning with the “Switch” strategy are two in three, double the odds of staying.”