[EM] a Borda-Condorcet relation

[Kathy Dopp]

Stephen,

I love it. Seems fair, precinct-summable, monotonic, fairly easy to
understand and count.

However, in the US incompletely ranked ballots with fewer rank choices than
the total number of candidates must be considered legal. Do you see any
problems with partial rankings or with US ballots that only allow voters to
rank their top three?

For clarification, does Borda give a rank of n to the 1st choice candidate,
n-1 to the 2nd choice, and so on?

Kathy

From: Stephen Turner <smturner0 at yahoo.es>
> To: election-methods at electorama.com
> Subject: [EM] a Borda-Condorcet relation
> Message-ID: <334970.4035.qm at web29705.mail.ird.yahoo.com>
> Content-Type: text/plain; charset="iso-8859-1"
>
> Dear EM fans!
> ?? ?? I was wondering if anyone can think of a source
> for the following simple observation, as it might make
> a nice little paper.? It amounts to seeing how the Borda
> count can be made Condorcet-compliant by
> replacing the mean by the median as detailed below.
>
> All ballots are total rankings of n candidates, and
> X, Y are two distinct candidates.? For any ballot,
> write s(X,Y) for (rank of Y) - (rank of X) which of
> course is the difference of their Borda scores.
>
> The s(X,Y) are all non-zero integers, at most
> n-1 in absolute value.
>
> We will suppose that there are no pairwise (i.e.
> Condorcet) ties.
>
> Given an election profile, write b(X,Y) for the mean
> of all the s(X,Y), and write c(X,Y) for the
> median of the same data set.
>
> Now b(X,Y)>0 iff X beats Y in the Borda count,
> and b(X,Y)>0 for all Y iff X is the Borda winner.
>
> Also c(X,Y)>0 iff X beats Y pairwise, and
> c(X,Y)>0 for all Y iff X is the Condorcet winner.
>
> What makes all this true is the fact that the mean of
> the differences (of the Borda scores for X and Y) equals
> the difference of their means; c(X,Y) is the median of the
> differences, which is not the same as the difference
> of the medians.
>
> If there is no Condorcet winner, then you
> could resort to one of the established methods
> to break the tie (RP, Schulze, minimax, ...)
>
> If there can be pairwise ties then c(X,Y)>0
> only implies that Y does not beat X pairwise.? There is a
> range 1<=c(X,Y)<=(n/2) - 1 in which either X beats
> Y or they tie - both are possible. ?
>
> As calculating the median is relatively expensive, the
> above probably is not useful as an algorithm.
>
> Any thoughts?
>
>
>
Kathy Dopp
http://electionmathematics.org
Town of Colonie, NY 12304
"One of the best ways to keep any conversation civil is to support the
discussion with true facts."

Fundamentals of Verifiable Elections
http://kathydopp.com/wordpress/?p=174

Realities Mar Instant Runoff Voting
http://electionmathematics.org/ucvAnalysis/US/RCV-IRV/InstantRunoffVotingFlaws.pdf

View some of my research on my SSRN Author page:
http://ssrn.com/author=1451051
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