# Using head-to-head results and Bayes' Theorem to modify predictions of sport/game contests that are initially derived from Elo-type ratings

I'm not even sure what the "probability of observing the data given the hypothesis" even means in this situation. In other Bayes' Theorem problems I've looked at, those probabilities are given or very simple to find. I thought I had some sort of answer, but after coding it up noticed that it was returning absurd results, specifically that a side with a 75% win chance as calculated from ratings and a 60% win rate in prior contests against the opposing side would win less than 60% of the time. Clearly there is something wrong with my calculations there. I was modeling the probabilities of the data given each side winning as though the data were being drawn from a binomial distribution with probability as implied by what the two sides' ratings would be updated to given that side winning. I also tried taking the binomial distribution's probabilities from ratings at the time of the previous contests, but the absurd results were still there, and it was far less obvious how to modify the binomial probability based on each side winning.

Where is the mistake in my approach here? How should I be finding the conditional probabilities that are reversed from the ones that I'm actually trying to calculate? Is this even not actually a Bayes' Theorem application because such conditional probabilities are not obvious, and I'm better off looking for the probability I'm most interested in a more direct manner?

Let's say your rating system says A consistently has a $60\%$ chance of winning against B. You have the following match data available: in four matches, the victors were A, A, B, A. In reality, the results of matches are not independent of each other (contestant confidence and other factors are in effect) but let's assume they are.
The prior odds of A winning were $3:2$. We multiply this by the odds of the four matches actually having those results given the prior odds, i.e. $3:2 \times 3:2 \times 3:2 \times 2:3 \times 3:2 = 162: 48 = 27:8$. This gives a normalized probability of A winning as $27/35 \approx 77\%$.