So, I thought I kind of understood Bayes and total probability, but I see now I am not clear on the calculation and which information to take in. I read an article that was using the Monty Hall problem, and it said that the proof was as follows:
Consider you chose door 1 and Monty opens door 3, which has a goat behind. Now, say A is the event that door 1 has a car behind and B that door 3 had a goat behind. So, in Bayes terms: $$P(A|B)=\frac{P(A\cap B)}{P(B|A)P(A)+P(B|A^c)p(A^c)}$$
My problem comes with the denominator. In the article they calulculate it as $\frac{1\times \frac{1}{3}}{1\times 1/3+1\times \frac{1}{3}}$That means he already uses the information that, because B is given, then all probabilites of the form P(B|__) are equal to 1. My doubt is that, if I can do this, then all calculations in total probability would be trivially one (we would have $P(\Omega )$.
I appreciate any help.