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 *P*_{0} of *I*-many fair coin flip measures on {*H*, *T*}. Let *U*_{X} be the subset of *Ω* where there are uncountably many *X*s (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 ℵ_{0} < *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

*P*

_{0}-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

*X*s (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

*P*

_{0}-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

*X*s, 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.

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