I can kind of understand the logic behind it, if you assume your door can't be affected by the probability of it, but the thing that still stumps me about this is how the probability for your door is "locked in."

You picked a door out of a set, and by opening any number of doors, the host has altered the set. The other door remaining went from being a 99/100 chance of having a goat behind it to being in a set of 98 knowns, and 2 unknowns. While the host can't choose it if it has a car, he also can't choose yours. You wind up with 2 identical doors and X number of open doors, with each door having a 50/50 chance given the re-evaluation.

I know this is supposed to be the wrong answer, but I can't see why it's wrong. If you have an explanation, I'd love to finally be able to understand this problem.

[–] [email protected] 9 points 1 year ago* (last edited 1 year ago) (1 children)

It's because the host has knowledge of the situation. Each door has 1% chance. You lock one door closed that the host can't touch, goat or car. The other rule for the host is they have to open all of the doors except one and they also can't open the car.

The door they leave is all possibilities from all doors at the begining minus your door (since the host couldn't mess with it), which is 1%. Your original choice determined that door would stay closed by the host so the host can't effect it's odds with their knowledge. There is only a 1% chance you locked the car behind that door picking randomly. Either you got the 1% and locked the car door or you didn't. The host will remove 98 goats without random chance and which will leave a car or a goat left. There's a 99% chance you made them leave the car with your first choice being 1% and a 1% chance you made them leave a goat. It's still the original odds, but it flips because the host has knowledge and either was forced to removed all possible bad choices and leave the car or you managed to hit the 1/100 chance at the start. There's a 99% chance you didn't, so switching has a 99% chance the host left it there.

[–] [email protected] 2 points 1 year ago

I think this might have been the answer that helped me the most. Most of all, it's that the Monty Hall problem isn't about you, it's almost entirely about the host's action of revealing doors.

There's a 98/99 chance he left that door because it's the car, or 1/99 because it's the goat (assuming the one left out of calculation is your door which he can't choose). Your original choice, whether or not you picked the car, is largely irrelevant. His actions can't affect your door because he can't choose it

You're not betting on a new set of 1/2, you're not even betting on the door itself having a new probability. You're betting on the act of the host revealing doors.

[–] affiliate 7 points 1 year ago (1 children)

that's a really good question. the probability getting locked in is very counterintuitive. i think it helps to think about what happens in each case.

let's say you pick one of the 99 doors with goats behind them. this happens 99% of the time. the host is tasked with opening 98 doors. of the 99 doors you didn't choose, one has a car behind them. the host does not have a choice about which doors to open, his hand is forced. in this sense, the 99 other doors are tied together: since you originally chose a door with a goat behind it, the host is forced to leave only the door with a car unopened. so, you switch and you get a car.

next, let's say you get lucky and pick the door with a car behind it. this happens 1% of the time. now the host gets a choice about which doors to open: he gets to pick one door with a goat to leave unopened. in this scenario switching gets you no car.

so, 99% of the time, switching gets you a car. i hope this is helpful!

[–] [email protected] 2 points 1 year ago

This definitely helped me look at it as a whole, and definitely started me down the right path of getting it. Thanks!