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James,<br>
Thanks for taking an interest in the comparison between your new "weighted
pairwise" (magnitudes) and my older "compressing ranks" ideas for completing
Condorcet. Unfortunately, when it comes to using high-resolution ratings
ballots to combine ranking and rating, these two are far from the only obvious
candidates.<br>
Here is another that is fairly simple and intuitive. High-resolution
ratings ballots. Inferring ranking from rating, eliminate the non-members<br>
of the Schwartz-set. Of the remaining candidates, each ballot approves
those candidates rated above average. Eliminate the candidate<br>
with the lowest approval-score. Begin again, and so on, until one candidate
remains.<br>
This and the compressing ranks method have in common that they are Condorcet
methods that try to rely on the ratings as little as possible.<br>
Your example:<br>
<pre>24: A 100 > B 1 > C 0
24: A 100 > C 1 > B 0
22: B 100 > C 99 > A 0
4: B 100 > C 1 > A 0
1: B 100 > A 1 > C 0
22: C 100 > B 99 > A 0
3: C 100 > A 1 > B 0
100 ballots. A>B>C>A, all 51-49.
</pre>
All candidates are in the Schwartz set, so the approval-scores are:<br>
<pre wrap="">A: 48
B: 49
C: 47</pre>
C has the lowest approval-score , so C is eliminated. Then A pairwise
beats B, so A wins. <br>
<br>
Regarding my "Condorcet completed by compressing ranks method", you wrote
(Sat.Jun.12):<br>
<blockquote type="cite">
<pre>Note: I would also suggest modifying the proposal such that the ranks are
not compressed unless there is no candidate not beaten by a majority.
(That is, I think that if there is a candidate who is beaten pairwise, but
only by a minority, and every other candidate is beaten by a majority, it
helps strategy-wise to elect that minority-beaten candidate without
further adjustments to the rankings.)</pre>
</blockquote>
CB: I hate this idea. In general I dislike "Winning Votes", but concede that
it "helps strategy-wise" in plain ranked-ballot Condorcet.<br>
I don't think it mixes well with ratings. I think that in the 3-candidate
cycle case, the fact that Compressing Ranks always picks the<br>
most approved top-cycle candidate and your Weighted Pairwise (Magnitudes)
method can pick the least approved definitely counts.<br>
Also your "Weighted Pairwise" method is more complicated and less intuitive.<br>
<br>
Yet another method I prefer would be "Approval Margins".<br>
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. Then measure the "defeat
strengths" by the differences in the candidates'<br>
approval scores. On that basis pick the Ranked Pairs (or maybe some other
pairwise method at least as good) winner.<br>
<br>
To take your example: <br>
A>B 48-49 = -1<br>
B>C 49-47 = +2<br>
C>A 47-48 = +1<br>
B wins.<br>
<br>
Chris Benham<br>
<br>
<br>
<br>
<br>
<br>
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