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

On your first choice you have a 33.3% of being right. But you also have a 66.6% of being wrong.

The opening of another door does not change the fact you only had a 33.3% of choosing the right door in the first place.

But with one door opened the full 66% chance of you being wrong “collapses” to the other closed door you didn’t choose.

Imagine there was 1,000,000,000 doors. You pick one. The host then opens up 999,999,998 doors. And asks if you want to switch. You clearly did not have a 50:50 chance of picking the right door even though there’s only two closed doors out of a billion. You had only a 1 in a billion chance of picking the correct door. And a 999,999,999:1billion odds you chose wrong.

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