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 still get confused by it, but the simplest way I can explain is that, when you pick a door, the chances you picked an empty door (no prize) is 2/3. The fact that the host reveals a door doesn’t change much. You still know there is a 2/3 chance that you picked the wrong door on your first choice and a 1/3 chance that you picked the right door. So when he asks if you want to switch, he’s really showing you that there were really always only 2 options: opening the 1 of 2 incorrect doors you most likely picked or picking the 1 door left that the host knows is correct.

_____________________________________________________________

|___ **your door** (which is most likely empty) ___|

______________________________________________________________

|___**the last unopened door** (that is most likely correct)___|

hope this helped

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