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
The Monty Hall problems come down to this, is it better to choose one door, or two doors? If you decide to stay with your original pick, you are basically saying that you want to keep your one door instead of the two other ones. When you change doors, you are picking both the other ones.
A lot of people get stumped by the door reveal, but that door reveal is actually 100% inconsequential. The prize can only be behind a single door, which means that most doors will be empty. In the two doors that you did not pick, one door *must* be empty. That does not improve or worsen your odds in any way, because it is expected that the door will be empty. Furthermore, the host knows not to open a door that has the prize, so the door will be empty.
Since this step is inconsequential, you can skip it. So, Monty says pick a single door. Then once you pick a single door, he skip opening an empty door and says “okay, do you want to keep your single door, or switch to the other two doors?” Which do you choose, one door or two?
Another way to help this problem is add the amount of doors. Let’s say that there was 100 doors, and you have to pick one. You pick one, and then Monty says “do you want to switch to the other 99.” Keep in mind that with those 99 doors, at least 98 of them will be empty. If you see a bunch of empty doors, that should not be a surprise. Which do you chose?
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