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I know the problem is easier to visualize if you increase the number of doors. Let's say you start with 1000 doors, you choose one and the announcer opens 998 other doors with goats. In this way is evident you should switch because unless you were incredibly lucky to pick up the initial door with the prize between 1000, the other door will have it.
I now recall there was a numberphile with exactly that visualisation! It's a clever visual
It really is, it's how my probability class finally got me to understand why this solution is true.
This is so mind blowing to me, because I get what you're saying logically, but my gut still tells me it's a 50/50 chance.
But I think the reason it is true is because the other person didn't choose the other 998 doors randomly. So if you chose any of the other 998 doors, it would still be between the door you chose and the winner, other than the 1/1000 chance that you chose right at the beginning.
I don't find this more intuitive. It's still one or the other door.
This is the way to think about it.
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
Edit: seems like this was already answered by someone else, but I didn't see their comment due to federation delay. Sorry
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.
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", since the two ensembles are now of equal size, but only one ensemble shrank, and it was always at 2/3 odds of containing the prize.
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.
Same here, even after reading other explanations I don't see how the odds are anything other than 50/50.
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.
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? Why does door 1 have a fixed probability when door 2 doesn't?
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.
How do we even come up with such amazing problems right ? It's fascinating.
This is fantastic, thank you.