Suppose we have a countably infinite fair lottery, in John Norton’s sense of label independence: in other words, probabilities are not changed by any relabeling—i.e., any permutation—of tickets. In classical probability, it’s easy to generate a contradiction from the above assumptions, given the simple assumption that there is at least one set *A* of tickets that has a well-defined probability (i.e., that the probability that the winning ticket is from *A* is well-defined) and that has the property that both *A* and its complement are infinite. John Norton rejects classical probability in such cases, however.

So, here’s an interesting question: How weak are the probability theory assumptions we need to generate a contradiction from a label independent countably infinite lottery? Here is a collection that works:

The tickets are numbered with the set

*N*of natural numbers.If

*A*and*B*are easily describable subsets of the tickets that differ by an easily describable permutation of*N*, then they are equally probable.For every easily describable set

*A*of tickets, either*A*or its complement is (or both are) more likely than the empty set.If

*A*and*B*are disjoint and each is more likely than the empty set, then*A*∪*B*is more likely than*A*or is more likely than*B*.*Being at least as likely as*is reflexive and transitive.

Here, Axioms 3 and 4 are my rather weak replacement for finite additivity (together with an implicit assumption that easily describable sets have a well-defined probability). Axiom 2 is a weak version of label independence, restricted to *easily describable* relabeling. Axiom 5 is obvious, and the noteworthy thing is that totality is not assumed.

What do I mean by “easily describable”? I shall assume that sets are “easily describable” provided that they can be described by a modulo 4 condition: i.e., by saying what value(s) the members of the set have to have modulo 4 (e.g., “the evens”, “the odds” and “the evens not divisible by four” are all “easily describable”). And I shall assume that a permutation of *N* is “easily describable” provided that it can be described by giving a formula *f*_{i}(*x*) using integer addition, subtraction, multiplication and division for *i* = 0, 1, 2, 3 that specifies what happens to an input *x* that is equal to *i* modulo 4. (E.g., the permutation that swaps the evens and the odds is given by the formulas *f*_{2}(0)=*f*_{0}(*x*)=*x* + 1 and *f*_{3}(*x*)=*f*_{1}(*x*)=*x* − 1.)

**Proof:** Let *A* be the set of even numbers. By (3), *A* or *N* − *A* is more likely than the empty set. But *A* and *N* − *A* differ by an easily describable permutation (swap the evens with the odds). So, by (2) they are equally likely. So they are both more likely than the empty set. Let *B* be the subset of *A* consisting of the even numbers divisible by four and let *C* = *A* − *B* be the even numbers not divisible by 4. Then *B* and *C* differ by an easily describable permutation (leave the odd numbers unchanged; add two to the evens divisible by four; subtract two from the evens not divisible by four). Moreover, *A* and *B* differ by an easily (but less easily!) describable permutation. (Exercise!) So, *A*, *B* and *C* are all equally likely by (2). So they are all more likely than the empty set. So, *A* = *B* ∪ *C* is more likely than either *B* or *C* (or both) by (4). But this contradicts the fact that *A* is equally likely as *B* and *C*.

## 4 comments:

Do you need to be so specific about ‘easily describable’? All that seems to be needed are two sets P ⊂ Q such that P, Q\P and Qc (=complement of Q) are all infinite, and Q is more likely than the empty set. (To match to the post, take P = multiples of 4 and Q = even numbers.)

First, a lemma: for any sets A and B with A ⊂ B and A and B\A both infinite, A and B\A are equally likely (write as A ≈ B\A). Why? Because there is a permutation that switches A and B\A.

Applying the lemma, Q\P ≈ P ≈ Pc = Q\P ∪ Qc. But, by the lemma, Qc ≈ Q which is more likely than the empty set by assumption. Given your (4), this is a contradiction.

This sort of reasoning assumes all sets are comparable. Some problems for standard numerical probabilities are avoided by declaring some sets non-measurable. Maybe something similar could work for comparative probability.

Ian:

The name of my game here is to work with as simple premises as possible.

One might worry that a re-labeling that cannot be described by us doesn't really count as the sort of re-labeling under which probabilities should be preserved. So I want to restrict to describable re-labelings. But there are lots of paradoxes in view once one starts to talk about "describability". (E.g., "the first indescribable ordinal" -- oops, haven't I just described it?) So instead of talking about describability in general, I restrict to an unproblematic simpler version.

Note that the reasoning does not assume that all sets are comparable. Rather, it assumes that (easily describable) re-labeling preserves probability.

*Correction:* As weak premises as possible, not as simple. :-)

Yes, I had missed the point. I now see.

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