The probability is not about the doors, it's about who has the goat - which obviously doesn't change by removing false hiding places. Let's say I have 10 doors

The probability is not about the doors, it's about who has the goat - which obviously doesn't change by removing false hiding places. Let's say I have 10 doors, and put a goat behind 1 of them at random. You pick 1 door at random, leaving 9 for me. So, there is 9/10 chance that I still have the goat, and now 1/10 chance that you have it. Now, I have 9 potential hiding places for this goat. BUT, I am willing to communicate to you which door I put it behind (if I have it, which 9/10 times I do). I tell you this by removing all but one (i.e. 8) of my doors. So, there's 1/10 chance that you have the goat. There's 9/10 chance that I have the goat, and have just told you where it is. I then give you the option to choose my door instead of yours. Clearly, you should switch. The key is that I'm removing not "one door" but "all but one" of my doors. In this way, I find the 3-door case to be the most confusing, because the minimum amount of doors is removed. With 3 doors, I only remove one door - which doesn't seem that spectacular. If there were 100 doors, I would be removing 98 of my 99 doors - which I think is easier to understand by intuition. In fact, because of this your odds of winning-by-switching increase with the number of doors(see n-variant http://en.wikipedia.org/wiki/Monty_Hall_problem#N-doors http://en.wikipedia.org/wiki/Monty_Hall_problem#N-doors).