Here's a generalization of STAR inspired by my ponderings on utilitarian methods. Each voter provides a preference ranking, as well as indifference lotteries over all three-tuples. These indifference lotteries are of the type: "If only A, B, and C were running, then I would consider B winning with certainty to be equal in value to A winning with probability p, C winning with probability (1-p)". The example lottery here is consistent with either the ranking A>B>C or C>B>A. Now, the method class. It's parameterized by an internal cardinal method. For simplicity's sake, let this be Range here. Then the method goes like this: If there are only two candidates, the majority candidate wins. So this is a type-two method - it doesn't handle pizza elections. For three or more candidates, there's a function that takes a group of three candidates and returns the loser. It determines the loser by conducting a Range election with only those three candidates, and with each ballot being normalized. The loser of the resulting Range election is the loser that the function provides. To be more precise about the normalization, suppose that a voter voted A>B>C and considers a 70% chance of electing A, 30% chance of C to be equal to electing B with certainty. Then the ratings are chosen so that: - 0.7 times A's rating + 0.3 times C's rating = B's rating - A is rated 1 and C is rated 0 (using the full range of the ballot). In other words, A is rated 1, B is rated 0.7, and C is rated 0. In general, the "certain candidate" (the middle one in the ranking) is rated p, where p is the probability of electing A in the indifference lottery, since p*1 + (1-p)*0 = p.[1] Finally, let the potential winner set be the smallest set of candidates so that every Range election with two from the set as well as someone outside, makes the one outside lose. Run a conventional Condorcet method on the members of the potential winning set and the provided rankings to determine the final winner. Does this distinguish between awful and good centrists while only using lottery information? Yes. LCR with a good centrist: 40: L>C>R, p = 0.8 30: R>C>L, p = 0.8 20: C>R>L, p = 0.5 The internal Range election goes like this: L = 40*1 + 30*0 + 20*0 = 40 R = 40*0 + 30*1 + 20*0.5 = 50 C = 40*0.8 + 30*0.8 + 20*1 = 76 The loser is L, so the PW set is {C, R}, and the C vs R election is decided in favor of C. Bad centrist: 40: L>C>R, p = 0.1 30: R>C>L, p = 0.1 20: C>R>L, p = 0.5 L and R's scores are as above, and then C = 40 * 0.1 + 30 * 0.1 + 20*1 = 27 The loser is C, so the PW set is {L, R}, and the L vs R election is decided in favor of R. Now, there may still be an incentive to strategically misreport. In particular, it doesn't solve the three-candidate Burr dilemma. Forest suggested that Euclidean normalized cumulative voting would fix the problem because the optima would not be found at the corners of the polytope, i.e the optimal strategic vote doesn't rate everybody either max or min. But I'm not sure just how to do the normalization. Standard L2 normalization involves subtracting the mean and dividing by the standard deviation; but AFAIK, quadratic voting proposals constrain all weights to be positive. The former gives a constrained setup like: ||v||_2 = 1 pv_A + (1-p)v_C = v_B v_A >= v_B >= v_C for an A>B>C voter with probability p on the indifference lottery, and output rating values given by the vector v; but this still has free variables. The latter gives: ||v||_2 = 1 v_B = pv_A v_A >= v_B >= v_C nonnegative which gives the solution v_A = 1/sqrt(p^2+1), v_B = pv_A v_C = 0. -km [1] Note that a completely indifferent ballot can be ignored, since it would add the same amount to every candidate's score. Thus we don't actually need to specify what value such a ballot should have. Dealing with equality, e.g. A=B>C, is harder, and I haven't done that here, but in the limit, p->1 for an A=B>C ballot (indifferent between certainty that A gets elected and certainty that B gets elected) and p->0 for an A>B=C ballot. I would think.

Oops, I made some mistakes in my previous post... let's try to correct them. On 19.10.2021 14:10, Kristofer Munsterhjelm wrote: > Now, there may still be an incentive to strategically misreport. In > particular, it doesn't solve the three-candidate Burr dilemma. Forest > suggested that Euclidean normalized cumulative voting would fix the > problem because the optima would not be found at the corners of the > polytope, i.e the optimal strategic vote doesn't rate everybody either > max or min. It should solve the Burr dilemma with honest voting, now that I think about it: if the candidates are Excellent, Good, and Bad; and a majority ranks Bad least, then any normalization that sets the lowest ranked candidate to 0 will ensure that Bad is evicted from the potential winner set; and then the Excellent vs Good contest is decided by Condorcet. So this suggests the following vote guide for STAR with three candidates: "Give your favorite 100% strength, your loathed candidate 0% strength, and the in-between candidate a percentage X% so that you're indifferent between having that candidate elected for sure, and an X% chance of your favorite winning, (100-X)% chance of your despised candidate winning". Perhaps a little too verbose :-) > But I'm not sure just how to do the normalization. Standard L2 > normalization involves subtracting the mean and dividing by the standard > deviation; but AFAIK, quadratic voting proposals constrain all weights > to be positive. > > The former gives a constrained setup like: >     ||v||_2 = 1 >     pv_A + (1-p)v_C = v_B >     v_A >= v_B >= v_C > for an A>B>C voter with probability p on the indifference lottery, and > output rating values given by the vector v; but this still has free > variables. > > The latter gives: >     ||v||_2 = 1 >     v_B = pv_A >     v_A >= v_B >= v_C nonnegative That's not just nonnegative weights, but with the additional constraint that the despised candidate has weight 0, which I forgot to mention. This idea comes from that if you have to pay for quadratic voting tokens, then you probably wouldn't spend any on the candidate you like the least; at least not in the context of a type-two method. I know it's still somewhat shaky a justification. -km