Fair lottery over the integers

Why can no lottery give every whole number an equal chance of being drawn?

"Pick a whole number at random" is a phrase people use without hesitation, and it sounds like a description of something doable: a lottery over a hundred tickets is fair when each ticket has chance 1/100, and there is no obvious reason the idea should refuse to extend.

But it does refuse — absolutely, not merely as an engineering difficulty. What is worth chasing is which expectation has to give way, because the impossibility is not really about infinity. Something narrower is doing the work.

Write down what "fair lottery" demands. Every integer n gets the same probability, call it c. The probability of the whole set of outcomes is 1. And probabilities of disjoint events add up.

Only two possibilities exist for a single number c: it is zero, or it is positive. So take them in turn and see which survives.

Suppose c is positive — say 0.001. Then any 1001 integers together carry probability 1001 × 0.001 = 1.001. But they are a subset of all the integers, whose probability is 1, and a part cannot exceed the whole. Contradiction, and a cheap one: it needed only finite addition and the fact that probabilities of subsets do not exceed 1. In general, whatever positive c you name, taking any set of more than 1/c integers breaks it. Positive c is dead.

So c must be zero. Every single integer has probability exactly zero — which sounds tolerable, since events of probability zero are familiar enough. Now assemble the whole from the parts. The integers are a countable collection of disjoint single-number events, so their probabilities add: 0 + 0 + 0 + … = 0. And that total was required to be 1. Dead as well.

Both branches fail, so no such lottery exists. But now the interesting part — the two branches did not fail for the same reason, and only the second used anything strong. The first used ordinary finite arithmetic. The second used the rule that probability adds over countably infinite collections, not just finite ones. That rule is Kolmogorov's axiom of countable additivity, and it is the only load-bearing assumption in the whole proof.

Which suggests a test. If countable additivity is the culprit, a uniform distribution should become possible again once the outcome set is too large for that axiom to reach. Look at the uniform distribution on the interval from 0 to 1: every individual point has probability exactly zero, and yet the total is 1. Why is that not the same contradiction? Because the interval's points are uncountably many, so countable additivity never gets to make its demand — there is no such thing as summing uncountably many zeros to a required total.

So the obstruction is not infinity. It is countable infinity, precisely: the integers are large enough that every point must get zero and small enough that those zeros are obliged to add up. The continuum escapes by being bigger.

Three refinements, and the first repairs the impression that the argument has closed the subject.

Something uniform over the integers does exist — it simply is not a probability measure in Kolmogorov's sense. Natural density assigns to a set of integers the limiting fraction of the first N it occupies: the even numbers get 1/2, the multiples of ten get 1/10, any single number gets 0, and the whole set gets 1. That is uniform, translation-invariant and normalised. It is also finitely additive and demonstrably not countably additive, and you can watch it fail: the whole set is the union of its singletons, each of density 0, and 0 + 0 + … is not 1. Density is not broken by that; it was never claiming countable additivity. So the honest statement of the result is narrower than "no fair lottery on the integers" — it is "no countably additive one".

Second, whether that axiom deserves its place is genuinely contested. De Finetti argued for finitely additive probability precisely so that uniform distributions over countable sets could exist. The mainstream keeps countable additivity because so much depends on it — above all the ability to pass to limits. The sets {1..n} increase to all of ℕ, and countable additivity is what guarantees their probabilities converge to the probability of the union. Under natural density they do not: each has density 0 while the limit has density 1. Give up the axiom and you lose that continuity, which underpins almost every convergence theorem in probability.

Third, an easy misreading to head off: this is not a statement about very large numbers being unreachable in practice. The proof never mentions size. It would fail identically for a uniform lottery over the rationals in [0,1], which are bounded and dense and countable — countability alone decides it.

And what observation would refute the account? Exhibit a countably additive probability measure on the integers giving every singleton the same probability. The argument says any attempt must fail at one of two identifiable places, so a candidate can be checked directly: name the common value c, and either it is positive and you can list enough integers to exceed probability 1, or it is zero and the singletons cannot sum to 1. A construction surviving both checks would overturn the result.

• How finitely additive probability behaves in practice, and which classical theorems survive without countable additivity.

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