Here is an argument for the von Neumann – Morgenstern axiom of independence.
Consider these axioms for a preference structure ≾ on lotteries.
If L ≺ M and K ≺ N, then pL + (1−p)K ≾ pM + (1−p)N.
If M dominates L, then pL + (1−p)N ≾ pM + (1−p)N.
If A ≾ pM + (1−p)N′ for all N′ dominating N, then A ≾ pM + (1−p)N.
If L ≺ M, there is an M′ that dominates L but is such that M′ ≺ M.
If M dominates L, then L ≺ M.
Transitivity and completeness.
0 ⋅ L + 1 ⋅ N ∼ 1 ⋅ N + 0 ⋅ L ∼ N.
Now suppose that L ≺ M and 0 < p < 1. Let M′ dominate L but be such that M′ ≺ M by (4). Let N′ dominate N. Then pL + (1−p)N ≾ pM′ + (1−p)N by (2) and pM′ + (1−p)N ≾ pM + (1−p)N′ by (1). This is true for all N′ that dominates N, so pL + (1−p)N ≾ pM + (1−p)N by (3).
Now suppose that L ≾ M. Let M′ dominate M. Then L ≺ M′ by (5). By the above pL + (1−p)N ≾ pM′ + (1−p)N. This is true for all M′ dominating M, so by (3) we have pL + (1−p)N ≾ pM + (1−p)N. Hence we have independence for 0 < p < 1. And by (7) we get it for p = 0 and p = 1.
Enough mathematics. Now some philosophy. Can we say something in favor of the axioms? I think so. Axioms (5)–(7) are pretty standard fare. Axioms (3) and (4) are something like continuity axioms for the space of values. (I think axiom (4) actually follows from the other axioms.)
Axioms (1) and (2) are basically variants on independence. That’s where most of the philosophical work happens.
Axiom (2) is pretty plausible: it is a weak domination principle.
That leaves Axiom (1). I am thinking of it as a no-regret posit. For suppose the antecedent of (1) is true but the consequent is false, so by completeness pM + (1−p)N ≺ pL + (1−p)K. Suppose you chose pL + (1−p)K over pM + (1−p)N. Now imagine that the lottery is run in a step-wise fashion. First a coin that has probability p of heads is tossed to decide if the first (heads) or second (tails) option in the two complex lotteries is materialized, and then later M, N, L, K are resolved. If the coin is heads, then you now know you’re going to get L. But L ≺ M, so you regret your choice: it would have been much nicer to have gone for pM + (1−p)N. If the coin is tails, then you’re going to get K. But K ≺ N, so you regret your choice, too: it would have been much nicer to have gone for pM + (1−p)N. So you regret your initial choice no matter how the coin flip goes.
Moreover, if there are regrets, there is money to be made. Your opponent can offer to switch you to pM + (1−p)N for a small fee. And you’ll do it. So you have made a choice such that you will pay to undo it. That’s not rational.
So, we have good reason to accept Axiom (1).
This is a fairly convincing argument to me. A pity that the conclusion—the independence axiom—is false.
