Let Ω be a countably infinite set, and suppose that we have a fair lottery on Ω: one member is going to be picked "at random", with no biases. For a subset A of Ω, we can ask how likely it is that the random element is a member of A.
Now, the probabilistic question we asked is independent of any kind of structure on Ω. We have no biases in the choice of an element of Ω. Consequently, we would expect our probabilities to be permutation-invariant. There is only one set of permutation-invariant finitely-additive real-valued probabilities on Ω that assigns probabilities to all finite sets, namely the assignment that P(A)=0 if A is finite, P(A)=1 if A is cofinite (i.e., the complement Ω−A is finite) and P(A) is undefined if A is neither finite nor cofinite.
But suppose we are not satisfied with undefined probabilities in the interesting case where A is neither finite nor cofinite, and want probabilities defined for all subsets of Ω. Let's work with qualitative probabilities, i.e., a pair of relations < and ~ on subsets of Ω, representing being less likely than and being equally likely as. As minimal conditions, let us assume:
- < is transitive and irreflexive (A<A never happens)
- ~ is an equivalence relation
- if A~B, B<C and C~D, then A<D
- if A⊆B, then A~B or A<B
Since we want our probabilities to be order independent, we shall assume:
- if π is a permutation of Ω, then A~πA,
Now, up to permutation, the following are all the types of subsets of Ω:
- finite subset of cardinality n, for n a natural number (0,1,2,...)
- cofinite subset whose complement has cardinality n, for n a natural number
- subset that is neither finite nor cofinite.
There is no way of satisfying 1-7 with an additive probability ordering, i.e., one satisfying the de Finetti additivity axiom that if A and B, as well as A and C, are disjoint, then A∪B<A∪C if and only if B<C.
But for intuition's sake we can model this probability ordering with interval-valued probabilities as follows. Let a and u be two positive infinitesimals such that u/a is infinite. Then P(Fn)=[na,na], P(Cn)=[1−na,1−na] and P(U)=[u,1−u].
In any case, the resulting measure is counterintuitive. If Ω is the natural numbers, then all sets that are neither finite nor cofinite have equal probability by permutation equivalence. And so the probability of our random number being divisible by 100 is the same as the probability of its being divisible by 2.
Moreover, we get the following problem. Suppose that we have an infinite number of tickets, one per each of an infinite number of ticket holders. One ticket is picked at random, without disclosure of which ticket it is. Then every ticket holder tosses a fair and independent die. Let A be the set of tickets held by ticket holders who tossed 6 and let B be the set of tickets held by ticket holders who didn't toss 6.
Question: Is it likelier that the winning ticket is a member of B than that it is a member of A, i.e., is it likelier that the holder of the winning ticket tossed a non-six than that he or she tossed a six?
Answer 1: Yes. Suppose Fred is the holder of the winning ticket. Then the probability that Fred tossed non-six is 5/6 while the probability that he tossed six is 1/6.
Answer 2: No. We know for sure (i.e., with probability 1) that infinitely many ticket holders tossed six and that infinitely many did not. Thus, A is neither finite nor cofinite, and the same goes for B. Thus A~B: the two sets have equal probability.
Ooops! Intuitively, Answer 1 is the right one. But it's hard to escape the reasoning in favor of Answer 2. So what should we say? Maybe that probability breaks down in infinite lotteries. And if so, then we better not be in an infinite multiverse.