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 host never reveals a door which has the prize behind it. You already know from the start that two doors don’t have prizes, so when you pick one you already know at least one of the doors you don’t pick has no prize.

Which has a better chance of having the prize, 1 door or 2 doors? Obviously 2 doors. One of those 2 doors is always going to not have the prize but the host eliminated it by showing you. So by switching you in essence pick the 2 doors option.

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