Problems with conditioned probabilities are among the most difficult to grasp. A 90 minutes lecture is usually not enough to teach them sufficiently. Even one of my heroes, the great Martin Gardner, once got another variant ("one of my two children is a son") wrong on the pages of Sci. Am.

I have since long given up using intuition for such problems. I simply use Bayes' formula to arrive at the sometimes very surprising correct solution.

The verbal formulation of the problem is often underspecified (more formally, the "event space" is not stated unambiguously) and therefore, there is sometimes more than one correct solution. However, even if the problem is described in a way that has only one correct solution, often people don't see it.

Two easy extremes: (1) If Monty knows where the prize is and is always mean, and only gives you, after opening a door without the big prize, a new choice in the case your first choice happened to be correct, the probability of winning by switching is nil.
(2) If Monty knows where the prize is and is always benevolent, and only gives you, after opening a door without the big prize, a new choice in the case your first choice happened to be wrong, the probability of winning by switching is 1.

The two more common solutions, one of which is counterintuitive. (1) Monty knows where the prize is, and always (a) opens a door without the big prize and (b) gives you the choice to switch, then you should switch (counterintuitively), for the probability of winning then is 2/3.
(2) Monty himself doesn't know where the prize is and (a) always opens one of the two doors that were not your first choice and (b) in those cases in which the big prize was not behind the now open door always offers you the choice to switch, the probability of winning (whatever you do) is 50%, the intuitive solution.

Since you don't know what Monty knows you should switch unless you think he is mean, for even in the worst case, your probability of winning by switching does not increase.

Of course, in Monty's case, one can retrospectively find the correct solution the empirical way. Someone has done that and has looked at all available videos of the show. The clear result: the majority of the people has not switched, but in the majority of the shows switching would have been better. However, across all shows the winning probability for switchers was less than 2/3 (though of course higher than 50%) which clearly hints that sometimes Monty actually was mean spirited.

Learning by doing it repeatedly is not so successful, by the way, as someone has posted. After 50 repetitions (in the switching wins in 2/3 of the cases variant) still half of the participants in an experiment did not switch, though the difference between 2/3 and 1/2 should be obvious after so many repetitions.

I once found a way to make the majority of the participants switch in an experiment even at the first opportunity by increasing the number of "doors" to ten. Actually, the student in her diploma thesis used ten upside-down cups under only one of which was a prize. After the first choice, she opened eight of the ten cups but never the cup with the prize (the participants were fully informed about the procedure). Then she offered the participants to switch and most of them did (but only a tiny minority in the three cups control). The probability of winning for switchers in the ten cups condition was 9/10, of course. Though most of the participants did switch, when asked about the probability of winning by switching, even most of the switchers said stubbornly 50%, same as in the three cups control condition. (So much for intuition about why we do what we do.)

As I said above, most of the conditioned probability problems suffer from an insufficiently specified event space. A last example: If someone throws two dice in a dice box (you cannot look at the result) and then takes out one of the two dice from under the box showing a 6, what is the probability that the other dice also shows a 6? The intuitive response, 1/6, is only correct, if the other guy has not looked at the dices before taking one out from under the box, that is if the probability for taking a dice showing a 6 was not higher then chance.

If however, your opponent had a look under the dice box before taking the dice out that shows a 6 and will always show you that one of the two dices has a 6 whenever he can (like in a well known German game of dice, in which the double 6 is the second highest throw), then the probability that the other dice also shows a 6 is 1/11 (hint for those that do not use Bayes' theorem to find the solution: the probability of a double 6 is 1/36, the probability for any of the other five combination with one 6 is 1/18) . It pays to know that when playing this game.

Wolfgang