I suggest a simple solution, build a simultion of the original Monty Hall problem run it many times, then record the outcome. How often does a switch end in a prize and how often does it end in failure?

I side with the the many doors solution as I 'solved' it in my own mind that way. The inital chance of choosing the correct door is tiny, however after the bad doors are revealed your chance of now getting the right answer are far better. Throughout the entire sequence of events it appears that switching is the way to go. After all the entire sequence is what generates the outcome not just the final switch:

First choice = 1 in 3
Second choice 1 in 2 (remember the first choice was 1 in 3 this means when you switch you are trading a 'bad' choice of 1 in 3 for a better one of 1 in 2, you were less likely to choose the right door on your first choice, it's a way of looking at it.)

-Ian