But we are talking about Monty Hall after all. So I could be wrong :P
P(initial choice is correct) = 1/3
P(host shows goat) = 2/3
P(host shows goat | initial choice is correct) = 1
and vice versa P(host shows goat | initial choice is wrong) = 1/2
Applying Bayes:
P(initial choice is correct | host shows goat) = P(host shows goat | initial choice is correct) * P(initial choice is correct) / P(host shows goat) = 1 * 1/3 / (2/3) = 1/2
So, initial choice now has 1/2 odds for being right. The host revealing a goat gave us information on whether he was playing the P=1/2 or the P=1 game.
Intuitively applying this with the 100 door version: the host getting lucky and revealing 98 goats in a row makes it fairly likely that the initial door was correct all along and they didn't get super lucky avoiding the car 98 times, as there's a good chance the contestant helped them by hiding the car.
If you are interested in more details, Wikipedia lists this as the "Ignorant Monty" variant.
It doesn’t matter that the host meant to do that. Now that the information is revealed, the probabilities have changed.
Obviously if the host revealed the car, that would also affect the probabilities!