Try this:
Instead of thinking through the various outcomes, consider 2 different contestants who adopt opposing strategies: Messers Stick & Change. Under what circumstances will each win the car?
If Mr Stick is playing, he would only win if he choose correctly at the first time of asking - on average 1/3 of the time
If Mr Change is playing, then he wins the car if he chooses the wrong door first time around because Monty is forced to open the other goat-door leaving only the car for him to change to (as he always does) - on average 2/3 of the time.
Clearly better to be Mr Change.
For the avoidance of doubt, this is the original version of the game where Monty does know where the car is & deliberately avoids it (and the contestant understands this). Also anal, I know, but you absolutely have to be clear about who knows what and when to understand puzzles like these.
I think it's fairly obvious that where Monty chooses blindly & his door isn't opened before you re-choose, then it doesn't matter whether or not you change doors. This is similar to drawing lots: it doesn't matter which order everyone chooses.
If Monty chooses blindly and his door is opened before you get to change, then again it doesn't matter. Mr Stick is still going to win only when he chooses the right door to start with - 1/3 of the time.
Mr Change will win when he chooses an incorrect door to start with (prob = 2/3) AND Monty chooses the other goat's door (prob = 1/2). Overall probability = 1/3.
We can ignore the cases where Monty opens the car's door because either strategy has an equal probability of success (0). If one strategy were better than the other, it would be because it succeeds more often in the cases when Monty does not choose the car.