I'm trying to solve the following problem:
You are given two unfair coins. You flip both of them and ones comes up heads $\frac{2}{3}$ of the time while the other comes up heads $\frac{1}{3}$ of the time. Given you had a uniform prior on the bias before flipping, what is the probability that the first coin is more biased than the second coin?
Given that we have a uniform prior on the bias I had started naming
$A$: "it comes up heads" with $P(A)=\frac{1}{2}$ and then the event $B$: 'it comes up heads $\frac{2}{3}$ of times in the first coin and $\frac{1}{3}$ of times on the second coin', therefore $P(B|A)= \frac{2}{3} \times \frac{1}{3} = \frac{2}{9}$. At this point I would apply Bayes's Theorem and update the belief on $P(A)$ but I'm not too sure how $P(B)$ should be calculated in this context..