This is an attempt to make precise something I said in the comments to my previous post, and analyze it. Let T(p) be the claim that p is true. Work with epistemic probabilities. Assume the following three principles:
- p entails T(p)
- If p entails q, then P(p) is no greater than P(q)
Let OF be the claim that the future is open, so that no proposition about a future contingent is true (i.e., it either lacks truth value, or it is false, depending on the version of OF). Let q be the claim that tomorrow I will toss a quarter a thousand times and each time it will land tails. Let p be not-q. Intuitively, P(p) is more than 1-2-1000 (more, because the chance that I will both to do the tossing is very small). Now OF entails not-T(p) (by definition of OF). Hence, P(OF) is no bigger than P(not-T(p)) = 1-P(T(p)) by (3). But P(p) is no bigger than P(T(p)) (by (1) and (2)) and hence 1-P(T(p)) is no bigger than 1-P(p). Thus, P(OF) is no bigger than 1-P(p). But P(p) is more than 1-2-1000. Hence, P(OF) is less than 2-1000. Hence, OF is not worthy of belief.
But in writing this argument out rigorously, it became clear that there is a way in which it begs the question against those defenders of OF who deny excluded middle. For (3) implies a probabilified version of the axiom of double negation (the axiom of double negation is: not-not-p entails p), and double negation has to be denied by those who deny excluded middle. So the defender of OF can get out of the argument if she denies the axiom of double negation, and gives a propositional probability theory compatible with that denial. But to deny double negation is to go pretty far down the road of implausibility–all reductio arguments go down the drain at that point. Of course one might maintain double negation for non-future-tensed claims, but that’s ad hoc.