Algorithmic Asset Voting: Difference between revisions
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== Notes ==
It could be possible for a voter to submit a ballot indicating partial preferences in certain pairwise matchups i.e. that they'd want to only give 0.2 votes to help Candidate A be in the winner set rather than candidate B, but 1 full vote to help Candidate A and/or B be in the winner set over Candidate C. So for example, if they voted with a rated ballot A10>B8, A10>C0 and B10>C0. See the "connections to cardinal methods" section of the Condorcet methods page for more information. Note that this may be possible even in the multiwinner case. <blockquote>2-winner example:
11 A>B
5 B
10 C>B
11 D>E
5 E
10 F>E
Condorcet PR methods should probably pick (B, E) as the winner set, because both candidates in the set are the Condorcet winners within their respective Hare Quotas (the top 3 lines and bottom 3 lines), so if for example you tried to go with (A, E), the 10 totally unrepresented C voters could give B their votes, giving B 10+5=15 votes total to beat A's 11. But if say all voters indicate that B and E are only 10% better than less-preferred candidates, this is no longer the case (B and E actually become Condorcet losers in their quotas, because they get only 5 + 1 or 1.1 votes in their pairwise matchups against their quota competitors, who have 10 or 11 votes) and then (A, D) looks like a better Condorcet PR winner set, since C and F, the only viable competitors left, simply have fewer voters preferring them than A and D.</blockquote>Note that a voter could even be allowed to give less than a full vote to any candidate, even their 1st choice, in all pairwise matchups.
The use of the [[KP transform]] on rated ballots and then converting those Approval ballots to ranked ballots allows voters to submit, for example, a rated ballot A5 B4 C3, and have it treated either as 0.2 votes A>B, 0.2 votes B>C, and 0.4 votes A>C (score for preferred candidate minus score for less preferred candidate divided by max score yields the number of votes in each matchup) or 1 vote A>B, 1 vote B>C, 1 vote A>C.
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Note that in a pairwise match-up between two winner sets, a third winner set can actually emerge as the winner. This is because some voters may prefer some candidates from one set and some from the other. For example, in a matchup between a set with only the majority's preferred candidates and another set with mostly candidates with almost no support, but one candidate whom a quota prefer, the final winner will actually be a new winner set where most of the candidates are the majority's preferred candidates and the final candidate is the quota's preferred candidate.
The Smith Set in Algorithmic Asset PR:<blockquote>
8 D>A>B>C
7 B>C>A>D
5 C>B>A>D
8 D2>A2>B2>C2
7 B2>C2>A2>D2
5 C2>B2>A2>D2
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>
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