My previous missive under this heading suggested using the entropy of a lottery to judge how far it deviates from the ideal of full cooperation.
Since then, a little reflection and experimentation has led to a refinement of this idea that is more appropriate: Instead of selecting (from among the feasible lotteries) the one that minimizes the entropy, we should minimize the difference of the entropy and the expected log of the winner's rating (on the randomly drawn ballot that determines the winner). (This log must have the same base as the log in the entropy calculation.) For those just joining in, the "lottery" methods that we have in mind select the circled candidate on a randomly drawn ballot. Different lottery methods have different ways of defining the circling function c from the set of ballots to the set of candidates. They all have the property that the rating of c(b) is positive on ballot b. A feasible lottery L has the property that for every ballot b, the rating of circled choice c(b) is at least as large as the expected rating of the winner on ballot b under the lottery L. The Random Ballot lottery that circles the top rated candidate on each ballot is (in general) feasible. The subtrahend in the difference mentioned above is the reciprocal of the number of ballots multiplied by the sum of log(b(c(b))) over all ballots b, where b(c) is the rating given choice c by ballot b. Example 1: 25 A>C 25 B>C 25 A>D 25 B>D For simplicity lets assume that both C and D are rated at some constant R<1 on all of the ballots that give them positive ratings. Then both the {A,B} and the {C,D} lotteries are feasible and both have minimal entropy, which simplifies to log(2). But when we subtract the other term, we get log(2)-log(1) and log(2)-log(R) in the respective cases. Since 1>R, the first difference is smaller, so the first lottery is preferred. Example 2: 20 A>F 20 B>F 20 C>F 20 D>F 20 E>F Suppose that every ballot rates F at R > 1/5. Let's compare the Random Ballot lottery with the consensus lottery that elects F with 100% probability: The respective entropies are log(5) and log(1). When we subtract the expected logs of the winner's rating we get log(5)-log(1)=log(5) and log(1)-log(R)=log(1/R). Since R>1/5, log(1/R)<log(5), so the consensus lottery is preferred. Nifty? ---- Election-Methods mailing list - see http://electorama.com/em for list info