Adding my own explanation, because I think it clicks better for me (especially when I write it down):

1. Pick a door. You have a 66% chance of picking a wrong door, and a 33% of picking the right door.
2. Monty excludes a door with 100% certainty
3. IF you picked a wrong door, then there's a 100% chance the remaining door is correct (so the contingent probability is p(switch|picked wrong) = 100%), so the total chance of the remaining door being correct is p(switch|picked wrong)* p(picked wrong) = 66%.
4. IF you picked the right door, then Monty's reveal gives you no new information, because both the other doors were wrong, so p(switch|picked right) = 50%, which means that p(switch|picked right) * p(picked right) = 50% * 33% = 17%.
5. p(don't switch|picked wrong) * p(picked wrong) = 50% * 66% = 33% (because of the remaining doors including the one you picked, you have no more information)
6. p(don't switch|picked right) * p(picked right) = 50% * 33% = 17% (because both of the unpicked doors are wrong, Monty didn't give you more information)

So there's a strong benefit of switching (66% to 33%) if you picked wrong, and even odds of switching if you picked right (17% in both cases).

Please feel free to correct me if I'm wrong here.