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
I completely understand how the breakdown of the chances works but it still doesn’t click for me. Even imagining 100 doors. Okay, I pick 2 and he opens all but 66. I understand from a maths perspective, because he knows which door doesn’t have the prize, it’s 99/100 (I’m not sure why not 98?). But that doesn’t invalidate your original choice. Your door may have always had the prize or it didn’t; at the time of choosing all options were equally valid. The prize doesn’t move. If we had the door with the prize the host would be just leaving an arbitrary door unopened, but because of the how the ratio now splits thats now the more likely choice? Some people have described it as saying you are picking between 1 door and the 99 other doors, but that’s not the choice that’s offered to you; you’ve just had your original choice confirmed in that its definitely not one of those 98 other doors and it’s either in yours or the door the host leaves (which they know for sure does or doesn’t have the prize)
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