Suppose that during an infinite past a fair die was rolled every day, and that this game will end in a year. You know all the outcomes of the past rolls. Before each roll, you are asked whether you think the roll will come up six. If you answer correctly, you get a dollar. Otherwise, you lose a dollar.
There is an obvious strategy: Always guess "No." Then out of six rolls, on average, you will win five times and lose once, so you will on average make about 67 cents per roll. Here's a very reasonable claim:
- Guessing "no" is the optimal strategy for an agent that does not have foreknowledge of the future.
But it turns out that, given the Axiom of Choice, there is a strategy that beats this, a strategy guaranteed that you will win infinitely often and lose at most finitely often, and hence that gives you a long-run average of a dollar per roll, rather than the measly 67 cents of our above strategy. The strategy is to use a variant of the solution to the fourth hat puzzle here. For the technically minded reader I'll sketch the strategy below.
But (1) is obviously true: it's clear that whenever you are being asked to guess, you should say "no", and surely that's the best policy. So (1) is both true and false on the above assumptions (including the assumptions needed to make the alternate strategy go). And hence I think we should reject the possibility of knowing the outcomes of a backwards-infinite sequence of coin tosses. And the best way to do that is to embrace causal finitism: to deny that anything (say, your current knowledge) can depend on infinitely many events.
For the technically minded reader, here's the strategy. Consider the set of all backwards-infinite sequences of die rolls. Say two sequences are equivalent if they differ in only finitely many places. For any equivalence class E of sequences, choose a member f(E) (by the Axiom of Choice). Now whenever you're asked to make a guess, you already know all but finitely many of the items in the actual world's sequence of rolls. So you know which equivalence class E the actual world's sequence will fall into. So you guess according to f(E). And since the actual world's sequence differs from f(E) in only finitely many places, you're right all but finitely often.