NOTE: I noticed a lot of people had trouble understanding the problem, so I tried to rephrase and organize everything here.
Looks like the original post was subsequently refined to include more legible formulas, which precipitated more meaningful answers. SO! You can ignore this post.
It seemed like there was some confusion here, partially about the question and partially about the answer. I noticed that all the info was sorta disparate and floating around, so I decided to organize everything here.
Summary of Problem: Inder Gill tried two different Bayesian formulations of the Monte Hall problem and got two different answers. So, why was this the case?
InderGill was using this formula comparison as guide:
1) Here's @Inter Gill's first formula in markup
2) But compare to Formula 2, where InterGill got the wrong answer.
Weird! So, why was this formula wrong?!
On Reddit, u/BurkeyAcademy had a pretty good answer (https://www.reddit.com/r/statistics/comments/4kabyn/probability_monty_hall_problem_getting_different/)
Basically, InterGill's second formula was misinterpreted.
The denominator in the second formula was off!
Below, BurkeyAcademy reformulates Inter Gill's formula 1.
Using A to indicate "prize behind door 1" and Bi indicates "he opens doors 1,2, or 3".
Here's the numerator for Bayes formula, which is 1/3*1/2:
Here's the denominator, with the SUM formulation
Important part of the answer
P(A) is equal to the sum of all P(A) possibilities.
Basically, P(A) is equal to the probability of getting the prize if you opened door1 + the same probability as if you opened door 2 and the same for door 3.
Because Monte Hall has already opened door 1, that one is equal to 0, as you can see when we fill in the formula with numbers.
The main difference between the correct denominator formulation (below) and the incorrect formulation (above) is that InterGill multiplied one of the probabilities by 1 instead of 1/2!
All of those multipliers need to sum to 1, otherwise P(A) has a probability greater than 1!
We're essentially computing the remaining probability of the prize behind behind one of the doors.
There are two doors, right? Originally the probability of selecting the prize was 1/3, but we're updating that probability, because there are only 2 doors. So the denominator comes to: 1/3*1/2 + 1/3+1/2
Hope this helps!