One way to think about it is to imagine Monty picking randomly in either case, but then (internally and invisibly) correcting his pick.
Assuming, WLOG, that we choose door 1, that leaves us with 6 equally likely cases just before that final correction (or lack thereof):
A) Car 1, Monty 2
B) Car 1, Monty 3
C) Car 2, Monty 2
D) Car 2, Monty 3
E) Car 3, Monty 2
F) Car 3, Monty 3
If Monty doesn't correct in cases C and F, then when he shows us a goat behind (say) 2 then we learn we are in either A or E - it's 50/50. If Monty does correct himself, then we might have been in A or E or F.
Assuming, WLOG, that we choose door 1, that leaves us with 6 equally likely cases just before that final correction (or lack thereof):
If Monty doesn't correct in cases C and F, then when he shows us a goat behind (say) 2 then we learn we are in either A or E - it's 50/50. If Monty does correct himself, then we might have been in A or E or F.