# Intransitive Preferences You Can’t Pump

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.

• But ar­bi­trar­ily low prob­a­bil­ities need not ex­ist.

I be­lieve for prac­ti­cal pur­poses, “I (or you) buy a cheap lot­tery ticket, and if it’s the win­ning ticket, then you pay me \$1” is low enough.

• That’s a measly one in a billion. Why would you be­lieve that this is enough? Enough for what? I’m talk­ing about the prefer­ences of a for­eign agent. We don’t get to make our own rules about what the agent prefers, only the agent can de­cide that.

Re­gard­ing prac­ti­cal pur­poses, sure you could treat the agent as if it was in­differ­ent be­tween A, B and C. How­ever, given the bi­nary choice, it will choose A over B, ev­ery time. And if you offered to trade C to B, B to A and A to C, at no cost, then the agent would gladly walk the cy­cle any num­ber of times (if we can ig­nore the in­her­ent costs of trad­ing).