## Monday, November 21, 2022

### Dominance and countably infinite fair lotteries

Suppose we have a finitely-additive probability assignment p (perhaps real, perhaps hyperreal) for a countably infinite lottery with tickets 1, 2, ... in such a way that each ticket has infinitesimal probability (where zero counts as an infinitesimal). Now suppose we want to calculate the expected value or previsio EpU of any bounded wager U on the outcome of the lottery, where we think of the wager as assigning a value to each ticket, and the wager is bounded if there is a finite M such that |U(n)| < M for all n.

Here are plausible conditions on the expected value:

1. Dominance: If U1 < U2 everywhere, then EpU1 < EpU2.

2. Binary Wagers: If U is 0 outside A and c on A, then EpU = cP(A).

3. Disjoint Additivity: If U1 and U2 are wagers supported on disjoint events (i.e., there is no n such
that U1(n) and U2(n) are both non-zero), then Ep(U1+U2) = EpU1 + EpU2.

But we can’t. For suppose we have it. Let U(n) = 1/(2n). Fix a positive integer m. Let U1(n) be 2 for n ≤ m + 1 and 0 otherwise. Let U2(n) be 1/m for n > m + 1 and 0 for n ≤ m + 1. Then by Binary Wagers and by the fact that each ticket has infinitesimal probability, EpU1 is an infinitesimal α (since the probability of any finite set will be infinitesimal). By Binary Wagers and Dominance, EpU2 ≤ 1/(m+1). Thus by Disjoint Additivity, Ep(U1+U2) ≤ α + 1/(m+1) < 1/m. But U < U1 + U2 everywhere, so by Dominance we have EpU < 1/m. Since 0 < U everywhere, by Dominance and Binary Wagers we have 0 < EpU.

Thus, EpU is a non-zero infinitesimal β. But then β < U(n) for all n, and so by Binary Wagers and Dominance, β < EpU, a contradiction.

I think we should reject Dominance.