Attempts at Probabilistic Lob’s theorem

X = “Th(X)” Z = “not Th(Z) OR X” Th reasons: If not Th(Z) then Z. if Th(Z), then Th(Th(Z)) so Th knows it’s not the “not(Th(Z)” case. So Th(X), so X, so Z. Thus Z. This is an example of Th(Z), thus by Z, we have X.

X = “Pr(X) is defined and > 0.7” Z = “Pr(Z) not > 0.7 OR X” Pr reasons: If Pr(Z) not > 0.7 then Z. If Pr(Z) is defined and > 0.7 then Pr knows it’s not that case, so Pr(X) is defined and > 0.7. Thus X, thus Z. Thus Z. Pr, knowing this, will give Pr(Z) defined and > 0.7 (in fact = 1)l thus also X with prob 1. what’s wrong with this argument? Pr can assign infinitesimal probability mass to Pr(Z) = 0.68, then Z will be true while keeping Pr’s loss bounded.

X = “Pr(X) has over 90% of its mass over 0.7” Z = “Pr(Z) has at least 10% of its mass <0.7 OR X” Pr reasons: If Pr(Z) has at least 10% of its mass <0.7, then Z. If Pr(Z) has over 90% of its mass over 0.7 then in all those cases Pr … wait no, it would reason that in general Pr knows “Pr(Z) has at least 10% of its mass <0.7” is not the case, but it can’t know that because it doesn’t know its beliefs.

This is the key distinction. In logic, Th(X) ⇒ Th(Th(X)) even if not Th(X) does not imply Th(not Th(X)). In probability, neither can be assumed.