Two minor comments. First, the bitstrings that you use do not all correspond to worlds, since, for example, Con(PA+Con(PA)) implies Con(PA), as PA is a subtheory of PA+Con(PA). This can be fixed by instead using a tree of sentences that all diagonalize against themselves. Tsvi and I used a construction in this spirit in A limit-computable, self-reflective distribution, for example.
Second, I believe that weakening #2 in this post also cannot be satisfied by any constant distribution. To sketch my reasoning, a trader can try to buy a sequence of sentences ϕ1,ϕ1∧ϕ2,…, spending $$2^{-n}$ on the \(n\)th sentence \(\phi_1 \wedge \dots \wedge \phi_n\). It should choose the sequence of sentences so that \(\phi_1 \wedge \dots \wedge \phi_n\) has probability at most \(2^{-n}\), and then it will make an infinite amount of money if the sentences are simultaneously true.
The way to do this is to choose each ϕn from a list of all sentences. If at any point you notice that ϕ1∧⋯∧ϕn has too high a probability, then pick a new sentence for ϕn. We can sell all the conjunctions ϕ1∧⋯∧ϕk for k≥n and get back the original amount payed by hypothesis. Then, if we can keep using sharper continuous tests of the probabilities of the sentences ϕ1∧⋯∧ϕn over time, we will settle down to a sequence with the desired property.
In order to turn this sketch into a proof, we need to be more careful about how these things are to be made continuous, but it seems routine.
Two minor comments. First, the bitstrings that you use do not all correspond to worlds, since, for example, Con(PA+Con(PA)) implies Con(PA), as PA is a subtheory of PA+Con(PA). This can be fixed by instead using a tree of sentences that all diagonalize against themselves. Tsvi and I used a construction in this spirit in A limit-computable, self-reflective distribution, for example.
Second, I believe that weakening #2 in this post also cannot be satisfied by any constant distribution. To sketch my reasoning, a trader can try to buy a sequence of sentences ϕ1,ϕ1∧ϕ2,…, spending $$2^{-n}$ on the \(n\)th sentence \(\phi_1 \wedge \dots \wedge \phi_n\). It should choose the sequence of sentences so that \(\phi_1 \wedge \dots \wedge \phi_n\) has probability at most \(2^{-n}\), and then it will make an infinite amount of money if the sentences are simultaneously true.
The way to do this is to choose each ϕn from a list of all sentences. If at any point you notice that ϕ1∧⋯∧ϕn has too high a probability, then pick a new sentence for ϕn. We can sell all the conjunctions ϕ1∧⋯∧ϕk for k≥n and get back the original amount payed by hypothesis. Then, if we can keep using sharper continuous tests of the probabilities of the sentences ϕ1∧⋯∧ϕn over time, we will settle down to a sequence with the desired property.
In order to turn this sketch into a proof, we need to be more careful about how these things are to be made continuous, but it seems routine.