Let's view this as a zero-sum game between you and Nature. The game state at each step is a set of sentences, which is initially empty. At each step, we consider the lowest numbered sentence that is independent from the set. You choose a probability to assign to that sentence, then Nature chooses whether to add that sentence or its negation to the set. The outcome of the game is your Bayes score after infinitely many steps. Every coherent probability assignment corresponds to a strategy you can play in this game and vice versa, while every model corresponds to a play of the game by Nature.
You have proved that the game has a minimax value. Is it true that you can achieve that value by always choosing your move so that the two possible subgames after Nature's next move have equal minimax values? I think your conjecture would follow from that, because your Bayes score on a particular model is the limit of subgame values on the corresponding path through the game tree.
A proof could go something along these lines. Let's say your move will assign probability p to the next sentence. The value of the left subgame increases monotonically from negative infinity at p=0, and the value of the right subgame decreases monotonically to negative infinity at p=1. So there's a unique p where they cross over, and choosing any other p would be worse.
The above argument also seems to show that the assignment you found is the only coherent assignment that assigns equal finite Bayes scores to all models. The reason is that any other such assignment would correspond to a strategy that doesn't always keep the WCB of the left and right subgames equal. Then we can define two strategies for Nature, "always choose the subgame with the lower WCB" and "always choose the subgame with the higher WCB", which would give you different Bayes scores.
That leads to your second conjecture. If you start with a constant 1/2 assignment and apply the update move repeatedly, you get a sequence of assignments whose WCB increases. You can use your convergence trick to get a pointwise converging subsequence from that. If the limit is a coherent assignment, then we know that it assigns equal Bayes score to all models, so it must be the one you found. Otherwise we can apply the update move some more.
It's a bit unsatisfying that we had to pick a converging subsequence, but I think we can also prove that the whole sequence converges. Namely, if the sequence of probabilities for any sentence has more than one limit point, then we can do the above procedure with both of those limit points separately. Since we're guaranteed to reach the same result on both paths, that means both limit points must be the same.
Sorry about the handwaving, it's very likely that at least some of these arguments are wrong :-)
So this is wrong. Imagine that P (The WCB maximizer) does the same on all models except one, and does better on that one model. Restricting to any finite list of sentences will not change the worst case bayes score. (In fact, I can prove that the set of models with bayes score near WCB must be everywhere dense.)
You have said a lot of correct things that are important though.
First, If the first conjecture is true, then an alternate defintion of P is "The unique coherent probability assignment for which Bayes score is constant," which seems like a...
In this post, I propose an answer to the following question:
Given a consistent but incomplete theory, how should one choose a random model of that theory?
My proposal is rather simple. Just assign probabilities to sentences in such that if an adversary were to choose a model, your Worst Case Bayes Score is maximized. This assignment of probabilities represents a probability distribution on models, and choose randomly from this distribution. However, it will take some work to show that what I just described even makes sense. We need to show that Worst Case Bayes Score can be maximized, that such a maximum is unique, and that this assignment of probabilities to sentences represents an actual probability distribution. This post gives the necessary definitions, and proves these three facts.
Finally, I will show that any given probability assignment is coherent if and only if it is impossible to change the probability assignment in a way that simultaneously improves the Bayes Score by an amount bounded away from 0 in all models. This is nice because it gives us a measure of how far a probability assignment is from being coherent. Namely, we can define the "incoherence" of a probability assignment to be the supremum amount by which you can simultaneously improve the Bayes Score in all models. This could be a useful notion since we usually cannot compute a coherent probability assignment so in practice we need to work with incoherent probability assignments which approach a coherent one.
I wrote up all the definitions and proofs on my blog, and I do not want to go through the work of translating all of the latex code over here, so you will have to read the rest of the post there. Sorry. In case you do not care enough about this to read the formal definitions, let me just say that my definition of the "Bayes Score" of a probability assignment P with respect to a model M is the sum over all true sentences s of m(s)log(P(s)) plus the sum over all false sentences s of m(s)log(1-P(s)), where m is some fixed nowhere zero probability measure on all sentences. (e.g. m(s) is 1/2 to the number of bits needed to encode s)
I would be very grateful if anyone can come up with a proof that this probability distribution which maximizes Worst Case Bayes Score has the property that its Bayes Score is independent of the choice of what model we use to judge it. I believe it is true, but have not yet found a proof.