Call me stupid. But I think that it makes no difference to your chances of winning to switch or not to switch.
Suppose you flip a coin 99 times. It turns up heads every time. What is the probability that the coin will turn up heads for the 100th time? It's obviously 1/2, not 1/(2^100). It is wrong to assume that past events have anything to do with probability of the present case.
Similarly with the case of the door, the problem can be misleading because it would appear that the probability of getting the right door is still 1/3. But we are told in the problem that one door is eliminated. In every reading of the problem this happens. The present case as stated by the problem is this: we have chosen one of two doors, and one of them is the right one. Therefore, the probability of having chosen the correct door is 1/2, regardless of how many thousands of doors have already been eliminated.
You can switch if you want, it really makes no difference.