On 10/28/2011 2:21 PM, capologist wrote:
> > ...
> Not quite what I'm looking for. ...
> ...
> I'm looking for a deterministic method for generating a "picture"
> (partial ordering) of how the voters, in aggregate, feel about the
> preferability of the available options. (What we're doing at this
> stage is more akin to a poll than an election.) It seems to me that
> the A>(M1,M2)>B ordering does not reflect the voters' preferences as
> well as the A>M1>M2>B ordering.
I entered your example into the (free) VoteFair-ranking service at
VoteFair.org and here is the results page:
http://www.votefair.org/cgi-bin/votefairrank.cgi/votingid=10305-48109-09917
VoteFair popularity ranking is mathematically equivalent to the
Condorcet-Kemeny method.
The "Data and Calculation Summary" section lists the pairwise counts,
and this might be the "tool" you are looking for. These pairwise
counts, which are the same for both the Condorcet-Schulze and
Condorcet-Kemeny methods, show that M1 is preferred over M2 by two voters.
Richard Fobes
On 10/28/2011 2:21 PM, capologist wrote:
See section 5 of my paper:
Not quite what I'm looking for. That section describes a non-deterministic
method for generating a complete linear order.
I don't require a linear order. I'm OK with a partial ordering.
I'm looking for a deterministic method for generating a "picture" (partial ordering) of how
the voters, in aggregate, feel about the preferability of the available options. (What we're doing at
this stage is more akin to a poll than an election.) It seems to me that the A>(M1,M2)>B ordering
does not reflect the voters' preferences as well as the A>M1>M2>B ordering.
I'm open to the possibility that the Schulze method is the wrong tool for this
purpose.
I'm also open to the possibility that the Schulze method is the right tool for this purpose,
and is serving that purpose effectively in this scenario. That would imply that, in some
meaningful sense, A>(M1,M2)>B is at least as good or a better picture of the voters'
preferences than A>M1>M2>B. This is counterintuitive but perhaps it makes sense and I
don't yet understand why.
I think the latter is likely the case. M1 and M2 are beatpath tied. What's going on in
this example is that there is a beatpath of strength at least 2 (using margins) from
every candidate to every candidate. Since M1's pairwise win over M2 is not stronger than
this value, it has no effect. Is this a case of a meaningful but weak signal being lost
in "noise"? Or is the strength-2 cycle itself a meaningful signal that, for
good if inscrutable reason, overrides the weak preference between the clones?
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