Am quite sure that the previous example would result in a 2/3 probability in favour of switching even if the host doesn't know. The thing is, the host effectively has one choice, because if he will always have another chance to choose the correct, empty door before the curtains are open, which is practically similar to prior knowledge (sort of...heh).

Thing is, assuming that he chooses the right door on his first try, then the situation is practically similar to him stumbling onto the right answer, sans curtain. The answer my friend suggested seems to imply that the different instances in which the i)host gets it correct on the first try, and ii) the host gets it correct on the second one, will give different odds for the player.

It may be well to argue that the knowledge for case ii) might change the odds.

In case i) the host is 2/3 certain that the other unopened door has the prize, whereas in case ii) the position of the prize door would be known to the host. However, I have yet to grapple with how this affects the player, since he could not possibly know that.

Am getting unnecessarily worked up.