3 ms·

Man, i still don't get it. Well, maybe i do? :shrug: The way i see it, the other door has a higher chance of being correct than the door you originally chose.

Man, i still don't get it. Well, maybe i do? :shrug: The way i see it, the other door has a higher chance of being correct than the door you originally chose. However the two are fundamentally different, are they not? In reality both doors are equally likely to be correct, no? So yes, changing would be better compared to your original answer - but "to change or not" evaluates both doors in my mind, and each is a 50% chance, no? Which i think is what you're saying here: > So your first choice is isn't any more likely to be correct that it was when you first you chose it. But nevertheless the problem has me a bit confused in what is expected to be "right". It seems to me there is no right, it is a 50/50, unless you are strictly answering the moot point of "Which is better? The door you first chose, or the potential other door?". Ie i guess i view the original choice as irrelevant the moment the variables change. I feel i misunderstand the goal of the "problem".

pessimizer 4y ago
> In reality both doors are equally likely to be correct, no? No, your original pick will always have a 1/3 chance of being correct. The door that gets opened will always have a 0/3 chance of being correct. The last door has to get all of the rest of the chance. ----- edit: I haven't tried explaining it this way before, but instead of one winning door and two bad doors, there's one door with a brand new car, one door with a giant mousetrap, and one door with a stuffed toy dinosaur. These are the ways it can go. A) You pick the door with the car. Monty either picks the dinosaur or the mousetrap, he doesn't really care. If you switch, you get the dinosaur or the mousetrap. B) You pick the door with the dinosaur. Monty is forced to pick the mousetrap. If you switch, you win the car. C) You pick the door with the mousetrap. Monty is forced to pick the dinosaur. If you switch, you win the car. ----- edit 2: your intuition is actually comparing this problem to other problems and not seeing the difference. Take the above scenario for example, but change it so Monty always reveals the dinosaur, no matter what. A) You pick the door with the car. Monty picks the dinosaur. If you switch, you get the mousetrap. B) You pick the door with the dinosaur. Monty picks the same door, and shows you that it is the dinosaur. Now you will definitely switch, and have an even chance to win the mousetrap or the car. C) You pick the door with the mousetrap. Monty picks the door with the dinosaur. If you switch, you will win the car. That's your 50/50.
- jwilk 4y ago
  This variant is even more confusing, because I'd rather get the stuffed dinosaur. :-P
EGPRC 4y ago
This only occurs because the host knows the locations of the contents and cannot reveal the player's door and neither which has the car. So the other that he leaves closed is always the best possible one that could be found among the rest, because if any of them has the prize, that is precisely which he will avoid to discard. If you notice, the switching door is equivalent to the selection that a second player that was cheating would make, a cheater that checked inside all the doors that the first did not pick, and took which preferred from them. In this way, it is obvious that the second player will win as long as the first starts failing, which occurs 99 out of 100 times in the 100-doors version. And the revealed doors are like which neither of them picked. So, when there are two doors remaining, you know that one is the first player's selection and the other is the cheater's selection, where the cheater was more likely to get the correct.
- lupire 4y ago
  It doesn't really matter if the host knows, as long as you are allowed to switch to an open door. If the host opens other door randomly, then 1/3 times the host accidentally shows the prize, and you still have 2/3 chance of winning by switching (measured from before the door was opened), switching either to the unknown closed door or the known winning open door. With 100 doors, switching gives you 1/100 * 0 + 99/100 * 1/99 * 1 + 99/100 * 98/99 * 1 = 1/100 + 98/100 = 99/100 But, measured from after the doors open, with host having opened other doors uniform randomly, in the case where you didn't already win by directly being shown the prize (free win happens with frequency (doors-2)/doors, since you have to pick a loser and the host has to pick a loser to hide), then your chance of winning by switching is 1/2, making the naive answer correct (and for the correct reason)!
  - EGPRC 4y ago
    Yes, you could still win 2/3 of the time by switching in all games, only that with those rules the probability for each remaining door once a goat is revealed would be 1/2, as you stated. So, you could also win 2/3 of the time by staying with your selection everytime a goat is shown and only switching to the car when it is shown. But what I think that confuses most people is why one option has more probabilities than the other when there are two remaining doors and the car must be behind one of them, which does not occur with those conditions. It is needed that the host did it deliberately avoiding to reveal the car.

pessimizer 4y ago

> In reality both doors are equally likely to be correct, no? No, your original pick will always have a 1/3 chance of being correct. The door that gets opened will always have a 0/3 chance of being correct. The last door has to get all of the rest of the chance. ----- edit: I haven't tried explaining it this way before, but instead of one winning door and two bad doors, there's one door with a brand new car, one door with a giant mousetrap, and one door with a stuffed toy dinosaur. These are the ways it can go. A) You pick the door with the car. Monty either picks the dinosaur or the mousetrap, he doesn't really care. If you switch, you get the dinosaur or the mousetrap. B) You pick the door with the dinosaur. Monty is forced to pick the mousetrap. If you switch, you win the car. C) You pick the door with the mousetrap. Monty is forced to pick the dinosaur. If you switch, you win the car. ----- edit 2: your intuition is actually comparing this problem to other problems and not seeing the difference. Take the above scenario for example, but change it so Monty always reveals the dinosaur, no matter what. A) You pick the door with the car. Monty picks the dinosaur. If you switch, you get the mousetrap. B) You pick the door with the dinosaur. Monty picks the same door, and shows you that it is the dinosaur. Now you will definitely switch, and have an even chance to win the mousetrap or the car. C) You pick the door with the mousetrap. Monty picks the door with the dinosaur. If you switch, you will win the car. That's your 50/50.

jwilk 4y ago
This variant is even more confusing, because I'd rather get the stuffed dinosaur. :-P

EGPRC 4y ago

This only occurs because the host knows the locations of the contents and cannot reveal the player's door and neither which has the car. So the other that he leaves closed is always the best possible one that could be found among the rest, because if any of them has the prize, that is precisely which he will avoid to discard. If you notice, the switching door is equivalent to the selection that a second player that was cheating would make, a cheater that checked inside all the doors that the first did not pick, and took which preferred from them. In this way, it is obvious that the second player will win as long as the first starts failing, which occurs 99 out of 100 times in the 100-doors version. And the revealed doors are like which neither of them picked. So, when there are two doors remaining, you know that one is the first player's selection and the other is the cheater's selection, where the cheater was more likely to get the correct.

lupire 4y ago
It doesn't really matter if the host knows, as long as you are allowed to switch to an open door. If the host opens other door randomly, then 1/3 times the host accidentally shows the prize, and you still have 2/3 chance of winning by switching (measured from before the door was opened), switching either to the unknown closed door or the known winning open door. With 100 doors, switching gives you 1/100 * 0 + 99/100 * 1/99 * 1 + 99/100 * 98/99 * 1 = 1/100 + 98/100 = 99/100 But, measured from after the doors open, with host having opened other doors uniform randomly, in the case where you didn't already win by directly being shown the prize (free win happens with frequency (doors-2)/doors, since you have to pick a loser and the host has to pick a loser to hide), then your chance of winning by switching is 1/2, making the naive answer correct (and for the correct reason)!
- EGPRC 4y ago
  Yes, you could still win 2/3 of the time by switching in all games, only that with those rules the probability for each remaining door once a goat is revealed would be 1/2, as you stated. So, you could also win 2/3 of the time by staying with your selection everytime a goat is shown and only switching to the car when it is shown. But what I think that confuses most people is why one option has more probabilities than the other when there are two remaining doors and the car must be behind one of them, which does not occur with those conditions. It is needed that the host did it deliberately avoiding to reveal the car.