But the issue is that by switching doors, you have a 66% chance of winning, it doesn't drop to 50% just because there are 2 doors, it's still 33% on the first door, 66% on the other doors (as a whole), for which we know one is not correct and won't choose.
I found the easiest way to think about it as if there are 10 doors, you choose 1, then 8 other doors are opened. Do you stay with your first choice, or the other remaining door? Or scale up to 100. Then you really see the advantage of swapping doors. You have a higher probability when choosing the last remaining door than of having correctly choosen the correct door the first time.
Edit:
More generically, it's set theory, where the initial set of doors is 1 and (n-1). In the end you are shown n-2 doors out of the second set, but the probability of having selected the correct door initially is 1/n. You can think of it as switching your choice to all of the initial (n-1) doors for a probability of (n-1)/n.
But the issue is that by switching doors, you have a 66% chance of winning, it doesn't drop to 50% just because there are 2 doors, it's still 33% on the first door, 66% on the other doors (as a whole), for which we know one is not correct and won't choose.