The big problem that makes it unintuitive is that people don't pick up the little piece of info that the game show guy knows whats behind the doors, and deliberately picks the goat.
Even if he doesn't know, once you get to the state where you have chosen a door and another door has been opened, revealing a goat, the same probability applies.
Yes, if the door revealed the car it would be different but that is not the scenario described. The problem specifies that the opened door revealed a goat.
Say the car is behind door number 1. Here are the possible scenarios (all equally likely - assuming you choose between the doors with equal probability and the host chooses between the remaining doors with equal probability)
If you want you can enumerate all of the outcomes for if the car was behind door number 2 and door number 3 but they will look the same, choosing the right door initially gives 2 possible (indistinguishable) outcomes. In both cases swapping loses. Choosing the wrong door also gives two possible outcomes. In one swapping becomes pointless, you've already lost but in the other swapping wins.
For simplicity, let's just say that whichever door the car is behind is door number 1, you just don't know how the doors are numbered when you choose.
So, if you choose door 1, swapping loses.
If you choose door 2 or 3 it is more complicated. You have a 50-50 chance of having a car revealed and swapping being pointless. However, the problem states that a goat was revealed so swapping wins.
You have a 1/3 chance of choosing door 1. This means you have a 1/3 chance of losing if you swap. You have a 2/3 chance of chosing door 2 or 3. This means that you have a 2/3 chance of winning if you swap.
Yes, before the door is opened, if you choose door 2 or 3 you have a 1/2 chance of losing immediately but that has already been resolved by the time you make the decision to swap.