The way I like to think of the Monty Hall problem is like this… if you had the choice of picking either one of the three doors or two of the three doors (if the car is behind either, you win it), you would obviously pick two of the doors to give yourself a 2⁄3 chance of winning. Similarly, if you had picked your original door and then Monty asked if you’d trade your one door for the other two doors (all sight unseen), it would again be obvious that you should make the trade. Now… when you make that trade, you know that at least one of the doors you’re getting in trade has a goat behind it (there’s only one car, you have two doors, so you have to have at least one goat). So, given that knowledge and the certainty that trading one door for two is the right move (statistically), would seeing the goat behind one of the doors you’re trading for before you make the trade change the wisdom of the trade? You KNOW that you’re getting at least one goat in either case. Most people who I’ve explained it to in this way seem to see that making the trade still makes sense (and is equivalent to making the trade in the original scenario).
I think the struggle is that people tend to dismiss the existance of the 3rd door once they see what’s behind it. It sort of drops out of the picture as a resolved thing and then the mind erroneously reformulates the situation with just the two remaining doors. The scary thing is that people are generally quite easily manipulated with these sorts of puzzles and there are plenty of circumstances (DNA evidence given during jury trials comes to mind) when the probabilities being presented are wildly misleading as the result of erroneously eliminating segments of the problem space because they are “known”.
Fail.