Let's analyze the classical problem. After you pick a door, one of the remaining two doors is opened.
With more doors (say, 11), it wouldn't make for an interesting show if you opened 9 more doors after the first one. A more reasonable approach would be to open one additional door and let the player choose to switch or stay with their choice each time. But that would only complicate the explanation, of course.
Let's dissect the 11-door case. We will do the following: the player picks a door, then the host a) opens one other door; b) opens 50% of the remaining doors; c) leaves one door closed.
IMO, the (a) case is the closest generalization of the original problem. You have a 1/11 chance to pick the car door on your first try. Then, one other door is opened, and if you switch to another random door, the best case probability of you getting the car is 10/11 * 1/9 = 10/99. This number is slightly bigger than 1/11.
In the (b) case you get a better result when switching, but to me that just looks like adding artificial bias. Obviously, the (c) case looks best because you've removed so much choice by leaving only one other door closed. Adding more doors does one thing: improves the probability of switching. That's good if this is what you're aiming for. But this turns it into a completely different problem.
So, you can see that the result depends on what you want to get. If you try to stay faithful to the original problem statement, adding more doors actually makes the benefit of switching less obvious.
In summary, I'm convinced that 3 is not an accidental number in the original problem statement. Whoever came up with it, they knew how to remove space for speculation by providing the only reasonable thing to do: open 1 door (or 50% of the doors) and leave 1 other door closed.