Probability for truly fair infinite lotteries

Long ago, in correspondence with Plantinga and me, Peter van Inwagen suggested that the only way to model a countably infinite fair lottery is by assigning probability zero to every finite set of tickets, probability one to every co-finite set of tickets (a subset A of a set B is co-finite [relative to B] provided that the set of members of B that are not in A is finite), and an undefined probability to every other subset.

Van Inwagen’s proposal has been growing on me. In a truly fair lottery, the ordering of ticket numbers is irrelevant. Therefore, if Ω is the set of tickets and π is any permutation of Ω, the probability of A should be the same as that of πA, with each defined if and only if the other is. In other words the probability function should be permutation-invariant.

Proposition. There are only two finitely-additive real-valued probabilities invariant under all permutations of a countably infinite set Ω: the trivial probability that assigns 0 to the empty set, 1 to Ω and is undefined for all other subsets, and van Inwagen’s probability that assigns 0 to every finite set, 1 to every co-finite set and is undefined for all other subsets.

There is something very appealing about van Inwagen’s proposal: it’s the only finitely-additive real-valued probability that really captures the idea of a countably infinite fair lottery. I can't remember if van Inwagen had the above proposition in the correspondence, but he might have.

Proof of Proposition: For any two subsets X and Y that are neither finite nor co-finite, there is a permutation of Ω mapping X onto Y. Thus, by permutation invariance, either all sets that are neither finite nor co-finite have a probability or none do. Suppose first that all do. In that case, they all have equal probability. Let A be the evens, B be the odds, C the numbers equal to 0 modulo 4 and D the numbers equal to 2 modulo 4. They all must have equal probability. But A is the disjoint union of C and D, so by finite additivity, if all three have equal probability, all three must have probability zero. And so does B. Thus, P(Ω) = P(A) + P(B) = 0, a contradiction.

So, only sets that are neither finite nor co-finite have a probability. If the only subsets that have a probability are ⌀ and Ω, we are done. Suppose some other subset has a probability. If that subset is co-finite, its complement will have to have a probability too, so in either case there is a finite non-empty subset A that has a probability. Let A′ be a finite non-empty subset that has the same cardinality as A but intersects A in only one element. By permutation invariance, A′ has a probability. Thus, so does the intersection of A and A′. Hence, at least one singleton has a probability. Hence by permutation invariance all singletons have a probability. By finite additivity, that probability must be zero. It follows that all finite sets have probability, and that probability is zero, and all co-finite sets have probability, and that probability is one.

Remark 1: Suppose that we allow the probabilities to take values in some non-Archimedean ordered field. Then there are more possibilities. Specifically, for any positive infinitesimal α, we can define a probability that assigns to every finite set the probability nα where n is the set’s cardinality and to every co-finite set the probability 1 − nα where n is the cardinality of the set’s complement. And these are the only extra possibilities.

Remark 2: If we drop the countability condition on Ω, and assume the Axiom of Choice, then in the setting of the Proposition we can prove that P(A) is 0 or 1 for every subset A for which P(A) is defined.