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

[Forest Simmons]

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