I have given some more thought on the problem of how best to complete Condorcet by using high-resolution ratings ballots,
and I've now concluded that the best is the method I defined in my last post (Tues.Jun.22):
"Approval Margins":
High-resolution ratings ballots. Inferring ranking from rating, eliminate the non-members of the Schwartz-set.
Of the remaining candidates, each ballot approves those candidates rated above average (and half-approves those rated precisely average).
Use the (inferred) rankings to determine the results of the pairwise comparisons between the remaining candidates.
Then measure the "defeat strengths" by the differences in the candidates' approval scores. On that basis pick the Ranked Pairs winner.
This could perhaps be "Automated-Approval Margins" (AAM) to distingush it from "(Plain) Approval Margins" which would use
ranked-ballots with an approval cutoff. Non-members of the Schwartz-set are eliminated. Then those ballots that either approve
all or none of the remaining candidates have their approval cutoffs moved the minimum distance so that this is no longer the case.
(In other words, those that showed approval for all the remaining candidates now approve all except the candidate/s they rank
last; and those that showed approval for none of the remaining candidates now approve the candidate/s they rank first.)
Then, as above, use the rankings to determine the results of the pairwise comparisons (what you would call the "directions of the
defeats") and the margins between tha candidates' approval scores to gauge the "strenghths" of the defeats, and on this basis use
RP to pick the winner.
(BTW, I don't consider this category of method, those that use ranked-ballots with an approval cutoff, to be nearly as important or
interesting as those that use high-resolution ratings ballots (with many more slots than candidates) or those that use plain ranked-ballots.)
So I no longer support Condorcet completed by Compressing Ranks, or Condorcet completed by Approval Elimination.
I think they are unneccessarily drastic. I scratched the Approval Elimination method when I discovered that it is vulnerable to
Pushover strategy. Take this example (from Adam Tarr) of sincere preferences:
49:R>C>L 12:C>R>L 12:C>L>R 27:L>C>R
C is the sincere CW. If the R voters Burry C, then normal (plain ranked-ballot) RP/BP/MM elect R.
If instead ratings are used to complete Condorcet, we might get this:
49:R100>L1>C0 06:C100>R99>L0 06:C100>R1>L0 06:C100>L1>R0 06:C100>L99>R0 27:L100>C99>R0
All candidates are in the Schwartz set, so each ballot approves the candidates rated above average, giving these approval scores:
R:55 C:51 L:33
Approval Elimination eliminates L, and then C pairwise beats R, and so C wins. (Approval and Compressing Ranks, equivalent in
the 3-candidate case, both elect R.) Both your Weighted Pairwise and my Approval Margins method elect C.
Suppose the R voters, as well as Bury (offensively order-reverse), insincerely raise their rating of L.
49:R100>L99>C0 06:C100>R99>L0 06:C100>R1>L0 06:C100>L1>R0 06:C100>L99>R0 27:L100>C99>R0
Now the approval scores are R55, C51, L82. Approval Elimination now eliminates C and elects R.
With our favoured methods, the R voters' strategy backfires and L wins.
I believe that AAM meets Minimal Defense/WDSC. If a majority prefer candidate y to candidate x, then they can simply prevent
the election of x by placing a sufficiciently large gap in their ratings somewhere between x and y, which they can do without any misrepresentation of their sincere rankings.
Adapting a Steve Eppley example: 46:X100>Z1>Y0 10:Y100>X1>Z0 10:Y100>Z99>X0 34:Z100>Y99>X0
Because of the X voters order-reversing, all the candidates are in the Schwartz set. The ballots approve the candidates they rate
above average, to give these approval-scores: X46, Y54, Z44.
The results of the pairwise comparisons (based purely on the inferred rankings) are X>Z, Z>Y, Y>X. Determining the "strengths"
of these defeats by the approval margins, we get:
X>Z 46-44 = +2
Z>Y 44-54 = -10
Y>X 54-46 = +8
RP locks Y>X and then X>Z, to give the final order Y>X>Z. Y wins.
If instead the X voters raise Z in their ratings:
46:X100>Z99>Y0 10:Y100>X1>Z0 10:Y100>Z99>X0 34:Z100>Y99>X0
the approval-scores become: X46, Y54, Z90.
X>Z 46-90 = -44
Z>Y 90-54 = +36
Y>X 54-46 = +8
RP locks Z>Y and then Y>X, to give the order Z>Y>X. Z wins, so the X voters' strategising has backfired.
Automated-Approval Margins (AAM), gives voters little or no incentive to exaggerate their ratings, and gets away from the nonsense
of voters wanting to rank differently candidates they rate the same, and so doesn't have the clumsy feature of the voters having to fill in
two ballots. Also it has been shown that Weighted Pairwise can elect the least approved candidate (in a 3-candidate election), and I
doubt that AAM can do that.
I hope that you (and others) found this interesting.
Chris Benham
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