To break the argument into pieces I understand, the set of times at which we flip the coin is a well ordered set, we can choose a strategy (if we grant the axiom of choice) that makes finitely many mistakes at n=1, and it’s obvious that if we have a strategy for n, the same strategy will work for n+1, so transfinite induction gives us that we can choose a strategy for any n.

What I don’t follow is how we make the jump from being able to choose a strategy for any n to being able to choose a strategy before an infinite number of mistakes has been made. It seems to me that this is like arguing that since for any n, the subset of the sequence 1/i up to and including 1/n has an upper bound, the whole sequence must have an upper bound.

So let me know if I’m missing anything.
Ohh, and hi, Kenny!

I agree that one must consider the case in which an infinite number of coin flips results in all heads. But that doesn’t mean that the probability of this situation is non-zero. You’re confusing probability 0 with impossibility, which are two distinct notions.

There can be no strategy to guess the next coin flip, the next coin will never care how the previous coins landed. If it did, it wouldn’t be a random flip.

The logical error is in assuming that a part cares about the whole.

“the requirement that the probability of an infinite sequence of heads be 0.”
The probability of a series of n flips having all heads approaches zero as n goes to infinity. Although the probability of an infinite sequence of heads goes to zero, one still must consider the case in which an infinite number of coin flips results in all heads. Think in terms of multiple universes. For the nth flip, the universe splits into two, one gets a heads and one a tails. But there is always a universe that has all heads. Extending n to infinity requires considering the universe with all heads. Ignoring this special case, or any other special number of heads or ratio of heads to tails, results in weird results.

If we want the measure to be more “Lebesgue-type” rather than “Borel-type”, then it’s probably a bit harder to show that the relevant events are unmeasurable, but it should still probably work out. Certainly the Borel-Cantelli argument will still work, because there is a constant positive lower bound on the probabilities.

I’m beginning to get a feel for the structure of the problem now (although it seems my intuitions are appalling.)

Actually, I was wondering how things would look if you didn’t do it with fair coins. Although translation invariance would fail, your finite permutation invariance would stick. But it seems odd that a lot of things might turn out differently if the coin wasn’t fair.

I think the argument can be generalized even to the case where you haven’t yet chosen the strategy. In this case of course the space is no longer just the set of sequences, but the set of pairs consisting of a sequence of flips together with a function for the strategy. I don’t know how we would distribute credences over the strategies, but no matter which distribution we use, if we do it in a way that’s independent of the distribution over the sequences of flips, then the same argument about translation invariance shows that the credence in p_n must still be 1/2 (if it’s measurable – but in this case it seems much more plausible that it must not be measurable, approximately for the reasons you suggest).

Anyway, given a strategy, we can again use translation invariance to find the joint distribution of a finite set of . For any set of k of these, consider all the conjunctions of the and their negations, and note that all of these events are just translations of the others, and thus must have the same probability, so the probability of any conjunction of k of them is just . In particular, this means that they are pairwise independent.

By the second Borel-Cantelli lemma, if there is a sequence of pairwise independent events, and the sum of their probabilities is infinite, then with probability 1, infinitely many of them occur. If there were an infinite set of that were all measurable, then by the above result, their negations would be pairwise independent and their probabilities would sum to infinity, and thus infinitely many of them would occur. But we have proven that only finitely many of them occur. Thus, there must not be infinitely many of them that are measurable.

Of course, I’d be much more convinced of this result if I could prove somehow more directly that one of these sets must be unmeasurable. But the reasoning so far all looks good, and it seems to entail that all but finitely many are unmeasurable, which is basically what I expected. This argument seems to work both before and after choosing a strategy.

If we assume a further sort of permutation invariance, namely that the probabilities are the same if you permute any finite set of indices (both in the sequence of flips and in the choices of representatives – I think this must be part of what you mean by “you don’t prefer any strategy”) then the fact that all but finitely many must be unmeasurable before you’ve chosen a strategy means that all of them must be unmeasurable before you’ve chosen a strategy.