I am trying to analyze a Monty Hall question with four doors (3 goats, 1 car) just so I can then apply the problem with n doors.

I applied Bayes' theorem, calculated the probabilities and am trying to do an expected value analysis. In order to do so, I am looking at every option a player of the game has but when I add up the probabilities, it gets a result greater than 1.

What I have so far (assuming only one goat is revealed):

You have a 1/4 chance of winning without switching, 3/8 if you win and switch, 3/8 if you lose after switching. I confirmed this answer with @Just_a_fool response here. However, shouldn't Just_a_fool also take in consideration the change of losing without switching (3/4 chance) in order to get every outcome? The problem with that is the probability adds up to 1.75. Am I missing something? Any help would be greatly appreciated!