The "Turkey" problem (Re: [EM] 2-rank and N-rank Condorcet)

[Rob Lanphier]

Very interesting.  I guess my criticism was based on not fully 
understanding N-rank Condorcet myself.  I've gone back and read the 
discussion, and I'm much more intrigued.

Rather than having a formula to determine the number of ranks, my sense 
is that N should be chosen based on what people are used to.  I'd 
recommend N=5, if only because it lends itself to this familiar format:

       Strongly                    Strongly
       disapprove     Neutral       approve
Joan   (1)     (2)      (3)     (4)     (5)
Jane   (1)     (2)      (3)     (4)     (5)
Jack   (1)     (2)      (3)     (4)     (5)

I have no idea what the theoretical implications of this particular 
choice would be.  I also don't know what the default option should be.  
Looking at it from a theoretical standpoint, I would prefer "0", but 
from a voter expectation perspective, I would almost assume that "3" 
would be the expected answer.  This question becomes particularly 
pressing when dealing with write-in candidates.

Rob

Forest Simmons wrote:

>On Fri, 25 Apr 2003, Rob Lanphier wrote:
>
>  
>
>>It seems like two-rank Condorcet makes things incredibly complex, and
>>creates more problems than it solves.
>>    
>>
>
>Two-rank Condorcet is nothing more or less than Approval; the beats-all
>winner is the Approval winner, so there is never any cycle to worry about.
>
>I believe that the N-rank Condorcet with N less than the number of
>candidates has the advantages that Kevin suggested as well as the obvious
>advantage of simplifying the ballot.
>
>In essence CR ballots with resolution N are used, but the ballots are
>scored pairwise, i.e. the ballot tallies are sums of pairwise matrices
>that keep track of how many ballots express preference of candidate i over
>candidate j.
>
>If N=4 the ballots could look like this ...
>
>
>Joan   (2) (1)
>
>Jane   (2) (1)
>
>Jack   (2) (1)
>
>etc.
>
>The ballot level of support for each candidate is the sum of the marked
>digits.  This sum would be a number between zero and three, inclusive.
>
>Kevin suggested using N=3.  In that case, the two's could be replaced with
>ones in the sample ballot above.  Then the sum would be a number between
>zero and two, inclusive.
>
>I suggest that the number of levels (i.e. the resolution) be on the order
>of the square root of the number of candidates, say the smallest integer
>whose square is no smaller than the number of candidates.
>
>If this rule were adopted, then there would be two levels for two to four
>candidates, three levels for five to nine candidates, etc.
>
>Forest
>
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