## Monday, January 14, 2013

### Fair infinite lotteries: What can we say?

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:

1. < is transitive and irreflexive (A<A never happens)
2. ~ is an equivalence relation
3. if A~B, B<C and C~D, then A<D
4. if AB, then A~B or A<B
5. ∅<Ω.
Say that AB provided A~B or A<B.

Since we want our probabilities to be order independent, we shall assume:

1. if π is a permutation of Ω, then AA,
where πA is the set of all π(x) for x in A.

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.
I.e., any two finite (or any two cofinite) sets of cardinality n are equivalent under permutation of Ω, and any two sets that are neither finite nor cofinite are likewise equivalent, and by (6) any permutation-equivalent sets have equal probability. Let Fn be any finite subset of cardinality n, let Cn be any cofinite subset whose complement has cardinality n, and let U be any subset that is neither finite nor cofinite. By 1-6, we then have:
1. ∅=F0F1F2≤...≤U≤...≤C2C1C0=Ω.

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 AB<AC 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.

#### 1 comment:

Dagmara Lizlovs said...

Alex:

Do you play any lotteries? I've played only one or two. That is when there is really some money up for grabs. I remember when once we had 100 Million or so on the winning ticket. One of my horsey friends and I discussed what we'd do if our tickets were the winning ticket. Both of us agreed that the very first thing we'd do is get a new pickup truck - something like an F-350 or a Dodge 3500, dualy, 4X4, 8 foot bed, crew cab, Cummins diesel. We couldn't think of anything else we'd do with the money except maybe get a 100 acre or so spread, maybe put a doublewide on it. Other than that, we think that lotteries are mostly a tax on the poor.