[EM] RE: [Condorcet] Can we come to consensus?

James Green-Armytage jarmyta at antioch-college.edu
Fri Sep 9 17:30:00 PDT 2005


>
>I don't see the connection between approval and consensus to be that 
>strong. 

	"Approval" and "consent" are synonyms.

>I think rankings can equally well be used to describe consensus 
>- in a different way. I'd say that electing e.g. a candidate that has 
>received no first place votes but is ranked second by most voters could 
>be called a consensus candidate. Now when I think about this, the 
>definition of Condorcet winner is actually quite close to defining a 
>"consensus winner" (based on available pairwise comparison 
>information).

	I disagree. I think that "consensus" has a meaning that is quite separate
from "majority rule". E.g. 55 voters: A>B; 45 voters B>A. A is a Condorcet
winner, but in my opinion not a consensus candidate.
	I think that the degree of consensus for a candidate is measured by the
number of people who view that candidate as acceptable. Hence I think that
approval voting comes closer to being a "consensus" method than any
Condorcet variant. When strategy is a factor in an approval election,
however, the concept of "approval" loses much of its original meaning;
hence I prefer Condorcet methods when broad consensus is unrealistic and
majority rule is the best realistic goal for the election.
>
>=> Unanimous majority in pairwise comparisons. I think it also has the 
>characteristics that it tends to elect "consensus candidates" like I 
>described above. I'm not a native English speaker, so I may not know 
>the use of word consensus in English language well enough, 

	Your usage is not unprecedented, but it is a minority usage and in my
opinion a source of confusion.

>We discussed MMC and Smith in March under title "majority rule, 
>mutinous pirates, and voter strategy". This time I asked only if Smith 
>should always hold (since that case is easier :-). I assume that your 
>earlier position is still valid and you want both MMC and Smith to be 
>always respected (tell me if not).

	I think that Condorcet methods should pass Smith, and therefore MMC as
well. As long as you're doing pairwise counts, it seems to me quite
unnecessary and inelegant to choose from outside of the top set.
>
my best,
James




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