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

But it’s not two doors, 50/50. You chose one of three so you have a 1/3 chance of having the right one and there’s a 2/3 chance of it being one of the other two. Now the host opens one of the other doors THAT THEY KNOW WON’T REVEAL THE PRIZE. That hasn’t affected the basic 1/3 to 2/3 split, nothing has changed as the host knows which door they can safely open. So that leaves the other door with the whole of the 2/3 chance.

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