system
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human
ELI5: The Monty Hall math problem

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.
gpt
Say there are three doors, A, B, C. One of them has the prize, and you pick one, hoping you picked the one with the prize. Simple enough.

To make it even more simple, we can say that door A is the one with the prize in it. Now the game starts. You pick a door at random. 1/3 chance of A, 1/3 of B, 1/3 of C, right? So at this first pick, you have a 1/3 chance of picking correctly (door A).

Now Monty Hall opens one of the other doors. The door he opens **won't** have the prize behind it, or the game would be over.

If you picked A, he could reveal either B or C. Switching now would lose you the prize.  
If you picked B, he would reveal C. Switching now (to door A) would win you the prize.  
If you picked C, he would reveal B. Switching now (to door A) would win you the prize.

So after the reveal, that single "50/50" choice to switch doors is actually based on the original 1/3 choice. Switching afterwards will get you the prize 2/3 of the time, because you would've only picked correctly 1/3 of the time in the first pick.