Monty Hall Problem

I’ve heard about the Monty Hall problem before but was reminded of it the other day when it was mentioned in an episode of Brooklyn Nine-Nine.  It’s based on a game show scenario where there are 3 doors.  Behind one door is a new car, the other doors contain goats.  You win if you guess right and pick the door with the car.  Three doors and no hints leave you with a 1 in 3 chance of winning (33%).  But the scenario gets complicated when the host, Monty Hall, opens one of the doors that you didn’t pick to reveal a goat and then asks if you want to switch doors.  Should you?

Keeping it simple, let’s run through the 3 possible scenarios when you pick Door #1 and stick with your choice:

1. The car is behind Door #1, Monty opens Door #2 (or #3) to reveal a goat, you WIN!
2. The car is behind Door #2, Monty opens Door #3, you lose
3. The car is behind Door #3, Monty opens Door #2, you lose

You win in 1 out of 3 scenarios above.  Now lets see what happens when you initially pick Door #1 but switch doors after Monty reveals one of the goats:

1. The car is behind Door #1, Monty reveals a goat behind Door #2, you switch to Door #3, you lose
2. The car is behind Door #2, Monty reveals a goat behind Door #3, you switch to Door #2, you WIN!
3. The car is behind Door #3, Monty reveals a goat behind Door #2, you switch to Door #3, you WIN!

When you switch doors after the goat reveal, you win in 2 out of 3 scenarios (66%).  So clearly it is better to switch.

There are many websites with more detailed, mathematical explanations of the above.  Betterexplained.com has a nice game where you can click through the scenario, picking a door and then choosing to stay or switch after a goat is revealed.  Stick with one strategy and see how it affects your odds of winning after multiple repetitions.