Let U = p(0,0)+p(0,1) = P(X=0), and V = p(0,0)+p(1,0) = P(Y=0). Both Bayesians took a Dirichlet, D(1,1,1,1), as the prior distribution of the p(i,j), and both computed the posterior distribution of U and V given that X and Y are independent, that is given p(0,0) = UV, or p(0,0)p(1,1)=p(0,1)p(1,0). The marginal distribution of both U and V is the beta distribution, Be(2,2) (density 6x(1-x) on (0,1)).
One Bayesian discovered a simple way to find the conditional distribution of U and V given that X and Y are independent. One way of generating the D(1,1,1,1) distribution is to choose U from the beta distribution, Be(2,2), and then choose P and Q are independent uniforms on (0,1), independent of U. Then, p(0,0) = UP, p(0,1) =U(1-P), p(1,0) = (1-U)Q, and p(1,1) = (1-U)(1-Q) have the D(1,1,1,1) distribution, and of course, V=UP+(1-U)Q. Here, P represents the conditional probability that Y is 0 given X = 0, and Q represents the conditional probability that Y is 0 given X = 1. It is clear that X and Y are independent if and only if P=Q.
In this form, it is a very easy matter to compute the conditional distribution of U and V given P=Q. First of all it is clear that U is not affected when one conditions on P=Q since U is independent of P and Q, and so U still has the Be(2,2) distribution. Moreover, when P=Q, then V is equal to this common value, and the conditional distribution of the common value of two independent uniforms given they are equal is still uniform. Thus, we have answered problem 1: The conditional distribution of U and V given X and Y are independent is that U and V are independent, U is Be(2,2) and V is uniform on (0,1).
Or have we? The answer is not symmetric in U and V! The other Bayesian modeled the joint distribution of the p(i,j) as above starting with V rather than U. He/she then arrived at the conditional distribution of U and V given X and Y are independent as that of independent U and V with V being Be(2,2) and U being uniform on (0,1). Of course it is best not to take sides when two Bayesians disagree. But can you help resolve the problem?