Well, there have been some interesting comments on my last post, but what I think we need to do, is review the situation.
In the Monty Hall problem, most people's reasoning would be: there are two remaining closed doors, so there is a 50% chance that the car is behind one or the other. Therefore, the your chances of winning the car are not improved by swapping doors, they stay the same.
In reality, there is still only a 1/3 chance that you picked the right door to start with, and a 2/3 chance that the car is behind the one remaining closed door. So you double your chances by swapping doors.
Many people still find that they cannot accept this argument. However, it becomes more obvious if there are more doors.
For example, suppose there are 10 doors. You choose one. The compere opens 8 of the other doors, revealing goats, leaving only two doors still closed, the one you chose, and one other door.
(BTW in this scenario it is to my mind self-evident that the compere knows which door the car is behind, otherwise there is a 4/5 chance that he will inadvertently open the door that has the car.)
Now, there is still only a 1 in 10 chance that you chose the right door. It follows that there is a 9 in 10 chance that the car is behind the one remaining closed door.
In that scenario, do you stick or swap? It's your call.