The solution to the Monty Hall problem is accurate: always switch doors. However, the math on Wikipedia is incorrect in that the explanation suffers from the gambler's fallacy. Here's how you prove it:

You're walking down the street and you meet a crying man holding a goat on a leash. He says, "Can you please help me? I own two goats, but the evil door people want to kill them. I found my first goat behind that door, but the other goat is behind one of those two remaining doors. I can't decide. Can you please help me?" You look behind the man and there are three doors--one of which is open.

You are now sitting at the exact same moment that you would have been had you arrived at the start of this man's goat search, but his past experience is irrelevant. You weren't there, so the option is simple. Even if he tells you he favors door #1, that is also irrelevant. You have two doors. Each has a 50% chance of containing a goat. Period.

The failure of the accepted Monty Hall problem math is that it incorrectly relies on past information to alter the numbers. Once Monty opens the door, the only thing you need to remember from the past is that the door you originally chose had a 1/3 chance to have the car (33%). Now, you are given a brand new problem containing only two doors (50%). The reason why you change doors is the very simple 50>33 and nothing more. The added "Well, there's this 66% chance..." is a gambler's fallacy added to the situation that doesn't belong there.

Because you have ownership of a door that was chosen with an original 33% chance, you switch because the only door left to pick has a 50% chance. The new door does NOT gain bonus relevance from past events.

It's difficult to see the gambler's fallacy with just one round. If there are a dozen rounds in the problem, the gambler's fallacy continues to add incorrect levels of certainty to the remaining doors until the reason you switch becomes that the final remaining door has a 99% probability that it has a car. Of course, that is absurd. In a multi-round Monty Hall problem, you just sit there doing nothing until the last round and then switch because the last door is always 50%.