Imagine the thing with 100 doors and imagine the host isn't opening any doors, but just telling you what's behind them and it becomes pretty obvious.
If there are only two choices remaining, I guess I don't understand the assertion that there is still a 1/3 chance of winning.
Suppose you were new to the contest and were allowed to put money down on the choice at the point that there were two doors left.
Then you repeated the bet every new contestant.
Would one door would be more profitable than the other?
That would be a completely different game. The goat doesn't get reshuffled. The doors that remain are not random and most importantly, you know which door you picked in the first round. It doesn't become a 50:50 chance just because there are two doors, as the goat isn't distributed over those two doors, but across all three.
Imagine 1000 doors. You pick one. In round two you are asked if you want to stay with that one door or pick all the other 999 doors at once. What do you do?
Case 2: Suppose there are n doors and Monty knows what is behind every one of the doors. You choose one door at random. Monty deliberately opens (n-2) of the remaining doors from the left to right, skipping the door with the car. Then he asks you if you want to switch. Should you switch or not?
Whether n=3 or n=100, it seems to me that it does not matter whether Monty Hall has complete knowledge or zero knowledge of the location of the car. You are required to make a choice under the condition where there is only one other unopened door and all the other doors did not reveal a car. The player's original choice was correct with probability 1/n and the probability of the complementary event must be (n-1)/n. The player's strategy of switching will result in a win with probability of (n-1)/n.