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

I think the easiest way to wrap your head around this is to increase the number of doors. Instead of 3, lets say there’s 100 doors. You choose a door, then the host opens 98 wrong answer doors. There’s only two doors left, but its of course not 50 50. Your original door had a 1 in 100 chance of being correct, so choosing the only other door left has a 99 in 100 chance.

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