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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 le
by 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 agoIt 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 agoYes, 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.