The Monty Hall math problem

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I was watching Brooklyn 99 Season 4 Episode 8 around the 5 minute mark

The problem goes “There are 3 doors behind one of which is a car. You pick a door and the host, who knows where the car is, opens a different door showing nothing behind it. He asks if you want to change your answer.

Apparently the math dictates that you have better chances if you change your decision. Why? 2 doors 50/50 chance, no?

One character (Kevin) says it’s 2/3 if you switch 1/3 if you don’t. What? How? Please help.

In: Mathematics

27 Answers

Anonymous 0 Comments

The explanation that clicked for me was to think of them as two groups. Group A is the door you choose. Group B is the other two.

After your first choice, there’s a 33% chance it’s in A and a 66% chance it’s in B.

Now what if they said, “Do you want to stick with A, or do you want to switch to B?” Of course you’d switch, because group B has two chances.

When they remove one of the bad ones, there’s still a 66% chance it’s in group B. By letting you switch after getting rid of one of the losers, they’re still giving you *all* of Group B.

It’s kind of like they let you pick two doors instead of one.

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