Suppose that we wish to model an infinite fair lottery with tickets numbered by integers by means of qualitative probabilities, i.e., a reflexive and transitive relation ≲ between sets of tickets that satisfies the non-negativity constraint that ∅ ≲ *A* for all *A* and the additivity constraint that *A* ≲ *B* iff *A* − *B* ≲ *B* − *A*. Suppose, further, that we want to have the regularity constraint that ∅ < *A* if *A* is not empty.

At this point, we want to ask what “fairness” is. One proposal is that fairness is strong translation invariance: if *A* is a set of integers and *n* + *A* is the set {*n* + *m* : *m* ∈ *A*} of all the members of *A* shifted over by *m*, then *A* and *n* + *A* are equally probable. Unfortunately, if we require strong translation invariance, then we violate the regularity constraint, since we will have to assign the same probability to the winning ticket being in {1, 2, 3, ...} as to the winning ticket being in {2, ...}, which (given additivity) violates the constraint that ∅ < {1}.

One possible option that I’ve been thinking about is is to require *weak* translation invariance. Weak translation invariance says that *A* ≲ *B* iff *n* + *A* ≲ *n* + *B*. Thus, a set might not have the same probability as a shift of itself, but comparisons between sets are not changed by shifts. I’ve spent a good chunk of the last week or two trying to figure out whether (given the other constraints) it is coherent to require weak translation invariance. Last night, Harry West gave an elegant affirmative proof on MathOverflow. So, yes, one can require weak translation invariance.

However, weak translation invariance does not capture the concept of fairness. Here is one reason why.

Say that a set *B* of integers is right-to-left (RTL) bigger than a set *A* of integers provided that there is an integer *n* such that:

*n* ∈ *B* but not *n* ∈ *A*, and

for every *m* > *n*, if *m* ∈ *A*, then *m* ∈ *B*.

RTL comparison of sets of integers thus always favors sets with larger integers. Thus, the set {2, 3} is RTL bigger than the infinite set {..., − 3, −2, −1, 0, 1, 3}, because the former set has 2 in it while the latter does not.

It looks to me that West’s proof straightforwardly adapts to show that that there is a weakly translation invariant qualitative probability that coheres with RTL ordering: if *B* is RTL bigger than *A*, then *B* is strictly more likely than *A*. But a probability comparison that coheres with RTL ordering is about as far from fairness as we can imagine: a bigger ticket number is always more likely than a smaller one, and indeed each ticket number is more likely to be the winner than the disjunction of all the smaller ticket numbers!

So, weak translation invariance doesn’t capture the concept of fairness.

Here is a natural suggestion. Let’s add to weak translation invariance the following constraint: any two tickets are equally likely.

I think—but here I need to check more details—that a variant of West’s proof again shows that this won’t do. Say that a set *B* of integers is right-skewed (RS) at least as big as a set *A* of integers provided that one or more of the following holds:

*A* is finite and *B* has at least as many members than *A*, or

*B* has infinitely many positive integers and *A* does not, or

*A* is a subset of *B*.

Intuitively, a probability ordering that coheres with RS ordering fails to be fair, because, for instance, it makes it more likely that the winning ticket will be, say, a power of two than that it be a negative number. But at the same time, a probability ordering that coheres with RS ordering makes all individual tickets be equally likely by (1).

To make this work with West’s proof, replace his *C*_{0} with the set of bounded functions that have a well-defined and non-negative sum or whose positive part has an infinite sum.