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I believe the confusion with "ignoring the car picked cases" comes due to thinking that they are additional games to the original ones, instead of being part of those that constitu…

That's not correct, because if you agree that one kind of result tends to happen more than the other, then it is easier that the single game that you are currently playing belongs …

The Monty Fall problem asks you about the case when the host has managed to reveal a goat. That only includes a subset of the total attempts that you would start selecting a door, …

Moreover, to understand why with a random revelation of the goat the chances of each option are 50%, you first need to understand the real reason why they are not 50% in standard M…

You can get that the answer must be 50% once the host just reveals a goat by chance using reductio ad absurdum. Notice that since the host is doing it randomly, it would be the sam…

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 …

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 alway…

In Monty Hall Game you improve your odds to be 2/3, not 1/2. If you think the staying door wins 1/3 of the time and the switching door 2/3, where is the car located in the 1/6 miss…