The asymmetry is that if you pick the one with the car first (only 1/3rd chance) then Monty Hall can show you either of the other doors. If you pick one without a car first (2/3rd chance) then Monty Hall must show you the door that doesn't have a car and the remaining door always has the car.

So 2/3rd of the time you have a 100% chance of a car if you switch, 1/3rd of the time you have 0% chance of a car if you switch. So switching is a 66.7% probability of a win and is the 'inverse' of if you picked the car right on the first guess or not.

The choice of which door Monty Hall shows you is constrained and isn't an independently random event since he never shows you the car when he opens a door.

You can also laboriously go through every event and "bifurcate the world" and then add up the probabilities, but you need to be careful about how you do it:

    1.  assign the car to a door (there are three possibilities)
    2.  have the contestant pick a door (there are now nine possibilities)
    3.  handle which door Monty Hall shows (this is tricky because for the three permutations in #2 where the contestant picked the car, then the door Monty Hall picks is unconstrained and so the world splits, but you are slicing a 11.1% chance in half for ~5.6%, while for the six permutations where the contestant didn't pick the car, Monty Hall is constrained so those remain 11.1% chances (at every level the probabilities have to add up to the probabilities before them and you can't violate 100% when you sum across all of them).
    4.  then have the contestant either switch or not switch and cut all your probabilities in half again.
    5.  now color everything according to winning or not and sum it all up.

I call that the "Quantum Many Worlds Interpretation Solution to the Monty Hall Problem" and if you do step #3 correctly you should get vos Savant's answer.