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sixes_and_sevens comments on Stupid Questions May 2015 - Less Wrong
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I get that you can do this in principle, but in the specific case of the Allais Paradox (and going off the Wikipedia setup and terminology), if someone prefers options 1B and 2A, what specific sequence of trades do you offer them? It seems like you'd give them 1A, then go 1A -> 1B -> (some transformation of 1B formally equivalent to 2B) -> 2A -> (some transformation of 2A formally equivalent to 1A') -> 1B' ->... in perpetuity, but what are the "(some transformation of [X] formally equivalent to [Y])" in this case?
You can stagger the bets and offer either a 1A -> 1B -> 1A circle or a 2B -> 2A -> 2B circle.
Suppose the bets are implemented in two stages. In stage 1 you have an 89% chance of the independent payoff ($1 million for bets 1A and 1B, nothing for bets 2A and 2B) and an 11% chance of moving to stage 2. In stage 2 you either get $1 million (for bets 1A and 2A) or a 10/11 chance of getting $5 million.
Then suppose someone prefers a 10/11 chance of 5 million (bet 3B) to a sure $1 million (bet 3A), prefers 2A to 2B, and currently has 2B in this staggered form. You do the following:
The net effect of those trades is that they still played gamble 2B but gave you a dollar or two. If they prefer 3A to 3B and 1B to 1A, you can do the same thing to get them to circle from 1A back to 1A. It's not the infinite cycle of losses you mention, but it is a guaranteed loss.