[EM] Re: completing Condorcet using ratings information

[Chris Benham]

  James,
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.
Here is another that  is fairly simple and intuitive. 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.  Eliminate the candidate
with the lowest approval-score. Begin again, and so on, until one 
candidate remains.
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.
Your example:

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.

All candidates are in the Schwartz set, so the approval-scores are:

A: 48
B: 49
C: 47

C has the lowest approval-score , so C is eliminated.  Then A  pairwise 
beats B, so A wins.

Regarding my "Condorcet completed by compressing ranks method", you 
wrote (Sat.Jun.12):

>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.)
>
CB: I hate this idea. In general I dislike "Winning Votes", but concede 
that it  "helps strategy-wise" in plain ranked-ballot Condorcet.
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
most approved top-cycle candidate and your Weighted Pairwise 
(Magnitudes) method can pick the least approved definitely counts.
Also your "Weighted Pairwise" method is more complicated and less intuitive.

Yet another method I prefer would be "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. 
Then  measure the  "defeat strengths" by the differences in the candidates'
approval scores. On that basis pick the  Ranked Pairs (or maybe some 
other pairwise method at least as good) winner.

To take your example: 
A>B 48-49 = -1
B>C 49-47 = +2
C>A 47-48 = +1
B wins.

Chris  Benham





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