The Monty Hall Problem is a famous statistical brain teaser that has caused never-ending debate over the past 30 or 40 years. It is based on the old TV gameshow “Let’s Make A Deal,” which was hosted by Monty Hall, and it goes something like this:

There are 3 doors. Behind 1 door is a fancy car, and behind the other two doors are goats. You first make a choice of one of the three doors. Then Monty Hall open up one of the remaining doors with a goat behind it and then asks you if you want to switch your guess to the remaining door or keep with your originally guessed door. So the question is: should you switch?

The answer is not very intuitive and sort of messes with most people’s basic understanding of statistics.

Here’s the quick answer: YOU SHOULD ALWAYS SWITCH!

Here’s the longer explanation:

The door you originally picked has a 1/3 chance of having the car both before and after Monty gets involved. When Monty picks one of the remaining doors to open with a goat behind it, that doesn’t change the statistics of the scenario. Crazy, right!?

So that means there is a 2/3 chance that the two doors you did not pick have the car. From those two, Monty eliminates a bad door. The remaining unpicked door still has a 2/3 chance of having the car, even though you are staring at just two unopened doors. That’s higher than 1/3, so you switch.

Most people forget the the host knows where the goats are. If you pick door 1, you have a 33% chance of getting it correct and a 66% chance of being wrong. Essentially what you’re being offered at the 2nd part is that if there is a car behind either of the doors you didn’t pick you win. Which is 66%.

Here’s the 3 possible scenarios that can be played to make it even more clear:
You pick door 1. The car is behind door 1. The host opens a door with a goat behind it (either door 2 or 3, since they both have goats).

You switch, you lose.
You stay put,you win.

2)

You pick door 1. The car is behind door 2. The host opens a door with a goat behind it (thus has to be door 3, since the car is behind 2, and you’ve chosen door 1)

You switch, you win.
You stay put, you lose.

You pick door 1. The car is behind door 3. The host opens a door with a goat behind it (thus has to be door 2, since the car is behind 3, and you’ve chosen door 1)

You switch, you win.
You stay put, you lose.

Switching = 66% success rate
Sticking = 33% success rate

If you are still not convinced, you can play around the the odds at this online simulation

http://www.montyhallsimulation.com/

 

-RSB