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Coscott comments on Even Odds - Less Wrong
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Here is another attempt to present the same algorithm, with the goal of making it easier to memorize:
"Each puts in the square of their surprise, then swap."
To spell this out, I predict that some event will happen with probability 0.1, you say it is 0.25. When it happens, I am 0.9 surprised and you are only 0.75 surprised. So I put down (0.9)^2 * D, you put down (0.75)^2 * D, and we swap our piles of money. Since I was more surprised, I come out the loser on the deal.
"Square of the surprise" is a quantity commonly used to measure the failure rate of predicative agents in machine learning; it is also known as Brier score. So we could describe this rule as "each bettor pays the other his or her Brier score." There was some discussion of the merits of various scoring systems in an earlier post of Coscott's.
I thought it was very interesting that my natural assumptions lead to a Brier score like system rather than Bayes score. I really don't think Bayesianists respect Brier score enough.
I thought it was interesting too. As far as I can tell, your result is special to the situation of two bettors and two events. The description I gave describes a betting method when there are more than two alternatives, and that method is strategy proof, but it is not fair, and I can't find a fair version of it.
I am really stumped about what to do when there are three people and a binary question. Naive approaches give no money to the person with the median opinion.
I wrote up an answer to this here http://bywayofcontradiction.com/?p=118
You could just do all three pairwise bets. That will not be fair, since not everyone participates in all bets. The middle man might just be guaranteed to make money though. (for some probabilities)