Monty knows what is behind each door. You pick first. We can all agree that when you do this you have one out of three odds to get the car, two out of three to pick the burger. Now follow this closely, this is the key: you can't pick both burgers; you only get one pick. If you picked the car, there are two burgers left. If you picked a burger, there is a car and a burger left. So regardless, there is still going to be a burger behind a door you did not pick. Remember, Monty knows this. Just for the sake of completeness, let's say that Monty, through some innate meanness, opens a door you didn't pick and the car is behind. Obviously you don't increase your odds by switching. You have a burger either way. But remember, Monty knows what is there, and to make the show more "exciting," he will probably open a door that hides a burger. If you picked a burger, he will open THE only door left that hides a burger. If you picked the car, he can open either door, and voila, there is a burger. But neither of these maneuvers increases the 1/3 odds that you picked right in the first place, it simply leaves you with 50/50 odds that EITHER THE DOOR YOU PICKED OR THE OTHER DOOR NOT OPENED BY MONTY HIDES THE CAR! You have 50/50 odds ON EITHER DOOR whether you switch or not. Think of it like this: the odds that car was behind one of the three doors is 1/3, and the odds that it is behind ONE of the other two is 2/3, but once you pick, you don't shift yourself to 2/3 on the car by switching your pick. You simply shift that initial 1/3 chance from one door to another. Got it?
No, wrong. If you choose a burger at first (2 out of 3 games), then Monty WILL choose the other burger, so the car WILL be behind the unopened door 2 out of 3 games. We are not shifting the initial 1/3 chance to the other door, we are leaving the 1/3 chance for a 2/3 chance since EVERY time we are wrong, Monty will essentially show us the right door.
Now, if Monty doesn't know where the car is and there is a 1/3 chance that he reveals it, then your odds depend on how the game is played at that point. If you just lose when this happens, then you will lose 1/3 by switching, lose 1/3 by Monty revealing the car, leaving an overall 1/3 win rate by switching. The same as if you don't switch.
If, instead, the game is void and you get to play again, THEN your overall chance of victory becomes 50/50. Here's how:
You play 36 games
12 games you choose the car at first, Monty reveals a burger, and you switch to the other burger and lose
24 games you choose a burger at first, 12 of those he reveals the other burger and you switch to the car, and the other 12 he reveals the car and the games are replayed
At this point your record is 12 wins, 12 losses. We play the 12 voided games again
4 of these you choose the car at first, he reveals a burger you switch and lose
8 of them you choose a burger first, 4 he reveals the other burger you switch and win, the other 4 he reveals a car and play again
Now your record is 16 wins, 16 losses, and we have 4 games to replay.
It should now be easy to see that the only way your overall odds of winning are 50% is if the game is replayed when Monty reveals the car, which would generally be a LOSS because it's as if he doesn't give you the option to switch at all. I think we can agree that a game will never be void and replayed, and if Monty doesn't know where the car is then your overall chance of winning will be 33%. 1/3 games Monty will reveal the car and you lose, and of the remaining 2/3 your odds will be 50/50, which of course 50% of 2/3 is 1/3, switch or no switch.
So, finally, if Monty is clueless you will win 1 out of 3 games whether you switch or not. If Monty knows where the car is, then you will win 1/3 by staying pat, and 2/3 by switching. The ONLY way to get your odds to 50/50 would be if games are replayed when Monty screws up, which would never happen.