I agonized over this. Recently, in fact. Here's the way it finally sank in to me:
The decision points are this:
1) The show puts the car behind one of the doors.
2) The player chooses a door.
3) Monty picks a wrong door to open.
4) The player decides whether to switch.
#1 and #2 are random. Given a long enough series of players, #4 is effectively random as well. The one that's not random is #3.
Monty cannot select either the door with the car or the door the player selected. If the player chose the door with the car, Monty's choice is a 50/50 shot between the other two. If the player didn't, there is only one door Monty can open. If the player guessed right, Monty has a 50/50 shot to pick a door, and the player has a stay-and-win or switch-and-lose choice. This only happens 1 in 3 times - the odds of the player guessing right the first time. If the player guessed wrong, Monty has no choice and the player has a switch-and-win or stay-and-lose choice.
The player standing there on stage has no indication of which scenario occurred. However, their guessing right only happens 1 time in 3. So odds are, they're in the switch-and-win.
Edit: Oh, and this assumes that Monty always opens a door. If Monty only opens a door some of the time, the player should probably assume psychological trickery and stay when he opens. That's just my conjecture.