Intransitive Preferences You Can't Pump

by zulupineapple 2mo9th Aug 20192 comments

2


In the usual argument of money-pumping we take an agent with preferences A>B, B>C and C>A. Then we offer it to exchange C+$1 for B, then B+$1 for A, and finally A+$1 for C. Now the agent paid $3 and ended up where it started.

The assumption here is that not only does this agent prefer A to B, it prefers A to B+$1. Of course, the price of $1 could be too steep, then we could look for something less valuable to exchange.

However, payments of arbitrarily low value need not exist! Sure, you can propose a lottery, where the agent only has to pay 1$ with probability p. But arbitrarily low probabilities need not exist. To offer a lottery, you need to have a physical method to generate events with that probability. If p equals 1 divided by Grahm's number, how many coins would you have to flip to run this lottery? Even if you said "the agent will pay $1 when a lump of solid gold materializes out of thin air", there are numbers lower than that probability. I hate to be an ultrafinitist, but it's true, extremely small (or large) numbers are not physically meaningful.

Note, here I am assuming the axiom of continuity, i.e. that if B<A<B+$1, then for some p, we must have B+p·$1<A. However, we should be able to violate this axiom as well.

What does this imply? Probably nothing. This perverse agent is almost indistinguishable from an agent which sets U(A)=U(B)=U(C). The rule is that you can violate any axioms, but only when it doesn't matter.