NCAA Tourney Second round probabilities

[Spektre]

At the start of the March Madness tournament, various sources fill out the entire tournament with probabilities for each team to win the various games. For example, you may find the probabilities for Gonzaga to be: Game 1: 99.5%, Game 2: 90%, Game 3: 78% and so on.

However, a number of these sources never update their probabilities after we know the first round’s results.

I am trying to understand how the second round of the March Madness tournament probabilities can be updated based on additional info from the first round.
Prior to the tournament start we have priorities of each team winning their game in Round 1 and Round 2.

P(T,N) will represent the probability of team T winning its Nth round game. Thus P(2,1) is the probability of Team 2 winning its first round game. In this example:
P(1,1) = 75, P(1,2) =36, P(2,1) = 25, P(2,2) = 7, P(3,1) = 81, P(3.2) = 50, P(4.1) = 19, P(4,2) = 7

Since I am only looking at the second round, the updated second round probabilities can only possibly be affected by the results of the four teams in the local bracket.

First round: Game 1-Team 1 plays Team 2, Game 2 -Team 3 plays Team 4
Second round: the Winner of Game 1 plays the winner of Game 2.
The first round games are played and Team 1 and Team 4 win.

Obviously: P(2,x) and P(3,x) are now 0, as those teams did not win in round 1 What are the new P(1,2) and P(4,2) ? Assume no other information as gathered other than the outcome of the first 2 round games,

I was hoping an application of Bayes' Theorem would work here. A - the event Team 1 wins Game 2. B - the event Team 4 wins Game 1. Then we are looking for P(A|B) = (P(B|A) * P(A))/P(B). We would have P(A) and P(B), but I could not figure out P(B|A) either..

[peacebyinches]

I like where your head is at (there definitely should be more focus on bayesian approaches in gambling) but I'm not sure if its necessary in this example. Once you have the matchup of Team 1 and Team 4 in the second round, the implied odds of P(1,2) and P(4,2) can be figured out from the vegas line (eg Team 1 has a moneyline of -300 which translates to around a 75% chance of winning). Theoretically the vegas odds to win the whole tournament can be derived from the implied odds of that team winning each round vs. each potential opponent they would face, weighted by their opponents chance of making it to that round.

So Kentucky at +800 to win it all includes 99% implied odds of winning round 1, then in round 2 (for example) 80% implied odds vs. the 7th seed and 85% odds vs. the 10th seed. Repeat this for all their possible opponents along with the implied odds for each opponent up until that round of the tournament and you'll probably end up with something in the ballpark of +800.

[Spektre]

Thanks for the response. Yes you can get the Vegas implied odds from their moneylines, but the point would be playing AGAINST the Vegas lines with the models used to create the original bracket probabilities.
In otherwords, these bracket odds can be used to make picks vs the Vegas lines for the first round...but unless they update the bracket following the first round, you do not get updated probabilities to play against the Vegas odds...unless that infomration can be cleaned from the original bracket and Bayes.