Algorithmic Asset Voting: Difference between revisions
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Some optional assumptions are:
* When a negotiator is indifferent between certain outcomes (i.e. because their voters equally ranked those outcomes), they use their assets to help pick the socially best of those outcomes. As an example, if the voters cast rated ballots <blockquote>49 A5 B4 3 A5 B5 48 B5</blockquote>then treat the 3 A=B votes as preferring B, because B has the most points, giving B 51 > A 49 votes, making B the winner even though more voters actually prefer A.
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Initializing an algorithmic Asset negotiation, D and D2 have the most votes (1st choices). But within each Hare Quota, ABC and A2 B2 C2 are in a Condorcet cycle. Therefore, once a few negotiating moves (pairwise comparisons) have been done, D and D2 won’t be in any of the 2-winner winner sets the negotiators cycle through. For example, if the negotiators are currently supporting (B, B2), and D and D2 attempt to gather enough support to win, 12 ballots prefer B to 8 for D or D2, and the same for B2. So the Smith Set here is all of the outcomes in the Condorcet cycles, which is a proper subset of all possible 2-winner sets.</blockquote>
Some reasoning for why this method is most likely Droop-proportional and even Hagenbach-Bischoff-proportional: mathematically, if any candidate has an HB quota, no matter how the votes are arranged among other candidates, the quota-preferred candidates can guarantee they are are or are tied to be one of the (number of winners) top candidates. Examples: with 1 winner, HB quota is 1/2 (half), and even if a non-quota candidate has all of the other 1/2 of the votes, the quota-preferred candidate is at least tied for being the top candidate. With 2 winners, HB quota is 1/3rd, and even if two non-quota candidates split the remaining 2/3rd of the votes perfectly evenly, the quota-preferred candidate is still tied to be one of the top two. And so on. It would appear that so long as you always elect from the Smith Set of winner sets, HB-proportionality is guaranteed, because winner sets that pass HB ought to always beat winner sets that don't. Consider that Minimax, a Condorcet method, fails Smith and happens to also fail mutual majority<ref>https://en.m.wikipedia.org/wiki/Mutual_majority_criterion#Minimax</ref>, and in the single-winner case mutual majority is equivalent to Droop proportionality.
== References ==
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