## Wednesday, March 16, 2022

### 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.

Aron Wall said...

But wouldn't requiring permutation symmetry would also mess up the (better behaved) case of selections of real numbers? There are permutations of the reals in the interval [0,1] which map measure 1 sets to measure 0 sets.

My own belief, as I mentioned earlier, is that uniform selection over countably infinite sets is simply impossible.

Alexander R Pruss said...

Yes, but that's precisely because the cases of a "uniform" selection of real numbers are cases of something that's not really a fair lottery, because the structure of the reals is relevant to the choice.

IanS said...

Norton gives essentially your argument here: J. D. Norton, “The Material Theory of Induction”. https://prism.ucalgary.ca/bitstream/handle/1880/114133/9781773852546_chapter13.pdf?sequence=16&isAllowed=y

Norton & Parker, “An Infinite Lottery Paradox” (google it, the url is very long), give two different responses. Norton takes your line. Parker suggests incomplete comparative probability based on set inclusion.

David Duffy said...

How does this tie into the expectation and d.f. of binomial random variables defined via draws from N, such as even(i), odd(i), mod(i,4)=0?

Alexander R Pruss said...

Set inclusion gives you what I call weak invariance under permutations (namely that A is less likely than B iff gA is less likely than B), but I think we want strong invariance (A is equally likely as gA).