I just wrote the following sentence in a draft of the (infamous) probability paper.

bq. On nonadditive theories, it is possible to get evidence for a disjunction without getting evidence for either disjunct, which seems odd.

Then I realised, on classical Bayesian theory it is possible to get evidence for a disjunction without getting evidence for either disjunct. At least I think it is. The following looks to me to be consistent.

Pr(p v q) = 0.6
Pr(p) = 0.4
Pr(q) = 0.4
Pr(p & q) = 0.2
Pr(p v q | e) = 0.64
Pr(p | e) = 0.4
Pr(q | e) = 0.4
Pr(p & q | e) = 0.16

In that case, e looks like it is evidence for p v q, but not evidence for either disjunct. But could there be a realistic case like that? Presumably there could – in this case we could just find out that p and q are probabilistically independent.

Of course what I should have written was that on these nonadditive approaches we can get evidence for an _exclusive_ disjunction without getting evidence for either disjunct. And that still seems somewhat odd. But it looks less forceful now that the claim is qualified.