The thing is, you pick the door totally randomly and since there are more goats, the chance to pick a goat is higher. That means there’s a 2/3 chance that the door you initially picked is a goat. The announcer picks the other goat with a 100% chance, which means the last remaining door most likely has the prize behind it
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Kissaki@feddit.de 11 months agoI don’t find this more intuitive. It’s still one or the other door.
moreeni@lemm.ee 11 months ago
SgtAStrawberry@lemmy.world 11 months ago
Don’t be sorry, your comment was the first time I actually understood how it works. Like I understand the numbers, but I still didn’t get the problem, even when increasing the amount of doors. It was your explanation that made it actually click.
Elderos@lemmings.world 11 months ago
I think the problem is worded specifically to hide the fact that you’re creating two set of doors by picking a door, and that shrinking a set actually make each individual door in that set more likely to have the prize.
Think of it this way : You have 4 doors, 2 blue doors and 2 red doors. I tell you that there is 50% chance of the prize to be in either a blue or a red door. Now I get to remove a red door that is confirmed to not have the prize. If you had to chose, would you pick a blue door or a red door? Seems obvious now that the remaining red door is somehow a safer pick. This is kind of what is happening in the initial problem, but since the second ensemble is bigger to begin with (the two doors you did not pick), it sort of trick you into ignoring the fact that the ensemble shrank and that it made the remaining door more “valuable”.
SnowmenMelt@lemmy.world 11 months ago
The odds you picked the correct door at the start is 1/1000, that means there’s a 999/1000 chance it’s in one of the other 999 doors. If the man opens 998 doors and leaves one left then that door has 999/1000 chance of having the prize.
UntouchedWagons@lemmy.ca 10 months ago
Same here, even after reading other explanations I don’t see how the odds are anything other than 50/50.
eran_morad@lemmy.world 10 months ago
read up on the law of total probability. prob(car is behind door #1) = 1/3. monty opens door #3, shows you a goat. prob(car behind door #1) = 1/3, unchanged from before. prob(car is behind door #2) + prob(car behind door #1) = 1. therefore, prob(car is behind door #2) = 2/3.
Kissaki@feddit.de 10 months ago
Following that cascade, didn’t you just change the probability of door 2? It was 1/3 like the other two. Then you opened door three. Why would door two be 2/3 now? Door 2 changes for no disclosed reason, but door 1 doesn’t?
eran_morad@lemmy.world 10 months ago
No, you didn’t change the prob of #2. Prob(car behind 2) + prob(car behind 3) = 2/3. Monty shows you that prob(car behind 3) = 0.
This can also be understood through conditional probabilities, if that’s easier for you.
crate_of_mice@lemmy.world 10 months ago
The point is, the odds don’t get recomputed after the other doors are opened. In effect you were offered two choices at the start: choose one door, or choose all of the other 999 doors.
Not_Alec_Baldwin@lemmy.world 10 months ago
This is the way to think about it.