A simple explanation on why to switch doors on the Monty Hall Problem

It is an underspecified problem.

The problem as I’ve seen it described is a one-time deal, you choose a door and then Monty opens a second door that has a goat. But the problem statement doesn’t clarify whether Monty must show a second door in all cases.

Let’s assume an evil Monty who knows where the car is, and gets any cars when you guess wrong. It is consistent with the problem as stated…you don’t know the motivation or information that Monty has from the problem statement.

If you guess right the first time, Monte is happy to show you the goat door. Because he knows you did a probabilistic analysis and he knows that when the goat door is opened, you will change your mind. And when you do, you always get the goat. Monty gets the car.

But when you choose the wrong door, let’s say Monty does not open a goat door. The problem statement doesn’t say that you know Monty will show a goat door when you make your choice every time. Evil Monty, in the case where you guessed wrong, just opens your original choice, which is not a car.

The problem as stated does not say that Monty always opens the goat door. Just that he did this one time. If you include Evil Monty scenarios, the game theoretic solution equilibrates to what you expect…makes no never mind whether you change your mind or not. It’s only when you think you can game the system that you leave yourself open to never winning.