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.