The reason it's important is that the problem, as usually phrased, is a conditional probability problem. Say you picked door number 1. Given that Monty opened door number 3 and it's empty, what's the probability you'll win by switching?

Now suppose that Monty will always open the empty door as advertised, but he will also always prefer to open a door with the smaller number, if he has a choice of two empty doors to open. If that is how Monty always acts, then given that he's opened door number 3, your probability to win by switching is 100%, and not the usual 2/3. If he's opened door number 2, then your probability to win by switching is 1/2, not the usual 2/3.

If Monty will always choose which of the two empty door to open by random, then the probability to win by switching is indeed 2/3.

There is a way to phrase the problem differently, so that it becomes an unconditional probability problem, and then it doesn't matter how Monty chooses the empty door to open. One way to enforce this interpretation is to phrase it as follows: Suppose some player always chooses to switch. Over many attempts, what will be the approximate proportion of his wins? The answer will be 2/3. But that's not how the problem is usually phrased.

Oh, wow, you're completely right. That connection eluded me; thank you!

Why has no one phrased the answer like that before? That makes way more sense then talking about "adding information". I never grokked the solution until you described it that way.

Your logic is correct but you are a making a different assumption, that Monty treats each door symmetrically. For example let's say Monty's strategy is to always open door #3 if he has a choice. Then if your choice is to switch to door #3 or to stay, you should choose "stay".