Uncountably many coin tosses and a technical problem for the many-minds interpretation

Suppose I toss infinitely many fair coins. By the Law of Large Numbers (and assuming an appropriate version of the Axiom of Choice), with probability one I will have infinitely many heads and infinitely many tails. But what if I toss an uncountable infinity of fair coins, and I want to know whether I will get uncountably many heads and uncountably many tails? Intuitively, surely I will. It would seem really unfair if I only got countably many heads and all but uncountably many were tails (or the other way around)!

But the usual mathematical model for this situation offers no such guarantee. Let I be an uncountable index set, let Ω = {H, T}^I be the space of coin toss outcomes indexed by I, and let P be the completion of the product P₀ of I-many fair coin flip measures on {H, T}. Let U_X be the subset of Ω where there are uncountably many Xs (where X = H or X = T). It can be proved (see Appendix) that U_X is saturated nonmeasurable, i.e., any measurable subset of it has zero probability and any measurable superset of it has probability one.

The same is true (and with the same proof) if we ask what the probability is that there is some specific cardinality κ of heads (or of tails), where ℵ₀ < κ ≤ ∥I∥. Again, we come up against a saturated nonmeasurable set of outcomes.

So what? Well, this leads to a mildly interesting technical problem for the Albert-Loewer many-minds interpretation of quantum mechanics. Albert and Loewer want their many-minds interpretation to allow for the supervenience of minds on the wavefunction. This requires that every branch of the multiverse always be populated by the same cardinality of minds. To ensure this, they prepopulate every branch with continuum-many minds, in the hope that every branching will keep continuum-many minds in every branch. But the above result shows that there is no guarantee of this. If continuum-many minds each, as it were, flip a coin whether to take branch H or branch T, we cannot even say that it’s likely that continuum-many will take each branch.

This might seem very interesting: it might seem to entirely undercut the physicalist supervenience behind their story. But that’s going too far. For it may be that the completion of the product measure on the coin-flip space {H, T}^I isn’t the right model. It may be that we should extend the measure to assign probability 1 to U_H and to U_T. This can, indeed, be mathematically done. (I haven’t checked to make sure that one can keep all the intuitive symmetries in the probability space, though.) Though in fact we can extend to assign other probabilities, too (though it then may be impossible to keep the symmetries).

Appendix: Sketch of proof that U_X is saturated measurable: Suppose that A is a non-empty P₀-measurable set. Then A is defined by a constraint on countably many of the factors in Ω. But no such constraint suffices can ensure that every member of A has uncountably many Xs (unless it ensures that A is empty), and hence it is not the case that A ⊆ U_X. It follows that the only subsets of U_X that are P-measurable have null measure. But for exactly the same kind of reason, a non-empty P₀-measurable set A cannot be a subset of the complement Ω − U_X either. For then the constraint on countably many factors would have to guarantee the lack of uncountably many Xs, while allowing A to be non-empty, and that’s impossible. It follows that the only supersets of U_X that are P-measurable have full measure.