[EM] Is "Inverse Nanson" better than standard Nanson?

[Forest Simmons]

Simpler than Nanson or Inverse Nanson but still satisfying the Condorcet
Criterion and making essential use of Borda is Borda Seeded Single
Elimination.

For a complete ranking use Borda Seeded Bubble Sort:

Rank the candidates according to the Borda Count, and then for each n
(from two to the number of candidates) allow the n_th ranked candidate to
percolate as far up the ranks as it can on the basis of pairwise
comparisons.

The pairwise matrix contains sufficient information for this method.

The incentive for insincere ranking is nil if not completely eliminated; 
why would you want to rank someone ahead of your favorite? The only way
your favorite would even have a chance of spoiling your compromise's
chances of winning would be if your favorite came out ahead of your
compromise in the Borda Count.  In that case minimal loyalty to your
favorite would say, "Go for it!" 

Since Borda (according to Rob LeGrand's simulations, for example) has
significantly higher expected social utility than any of the Condorcet
methods, we can be confident that Borda Seeded Single Elimination and
Borda Seeded Bubble Sort will be near maximal with respect to expected
social utility among methods that satisfy the Condorcet Criterion.

This method is simpler than Borda Completed Condorcet because it doesn't
require computing the Smith set.

Furthermore, Borda Completed Condorcet does NOT effectively eliminate the
incentive to rank Compromise above Favorite, because one wants to give
Compromise every chance in the event of completion, and Compromise is much
more likely to be in the Smith set than Favorite, etc.

I think that Borda Seeded Single Elimination would be much easier to sell
to the public than any of the Condorcet methods because the algorithm is
so simple and it isn't necessary to consider cycles. Every elimination is
done by majority decision, so the method satisfies "majority rule" to a
much greater degree than IRV, for example.

The average voter could understand the method as a type of runoff
simulation, similar to sports playoffs where the teams are seeded
according to their pre-playoff performance.

Borda Counts are used extensively for rankings in sports.

The Borda Count for a candidate divided by the number of voters is just
the average rank of the candidate, so the seeding is done according to
the average ranks of the candidates. No mention of "Borda Count"
is necessary.

Among methods based on pure preference ballots (which lack approval
information) this Borda Seeded Single Elimination might be hard to beat
for a public proposal.

Personally, I believe that Approval ballots, CR ballots, Dyadic ballots,
and preference ballots with approval cutoff information are all
superior to mere preference ballots, and that Five Slot Approval (FSA, so
christened by Joe Weinstein) is the best single winner public proposal so
far. 

Forest



 On Tue, 3 Jul 2001, Blake Cretney wrote:

> Richard Moore <rmoore4 at home.com> wrote:
> > As someone -- I forget who -- pointed out here some time
> > ago, you can calculate Borda results from a pairwise matrix.
> > Just add the figures in each column (or row, depending on
> > how you organize your matrix). This assumes the matrix
> > contains counts of all votes, not just winning votes. So,
> > it could be deemed a pairwise method.
> 
> In fact, that is how you would carry out Nanson in practice.  Once you
> have the pairwise matrix, you can easily see the Borda scores for any
> subset of the candidates.
> 
> On Sun, 01 Jul 2001 23:12:52 -0700
> Michael Rouse <mrouse at cdsnet.net> wrote:
> 
> > We can also invert Nanson's method like this:
> > 1. Calculate the Borda count for all candidate.
> > 2. Drop the highest-scoring candidate.
> > 3. Re-order the remaining candidates.
> > 4. As long as there are two or more candidates remaining, return to
> step 1.
> --snip--
> > 5. Put the final "loser" candidate in the appropriate position on a
> second 
> > list, and remove from the original list.
> > 6. Re-order the remaining candidates.
> > 7. As long as there are two or more candidates remaining on the
> original 
> > list, return to step 1.
> 
> Let me first suggest another way you could use Nanson's method to form
> a complete ranking.  Instead of assuming the first dropped candidate
> is ranked last, apply Nanson until you get a winner.  This winner is
> ranked first.  Remove this candidate from consideration, and apply
> Nanson to the remaining candidates to get the second candidate. 
> Remove it, apply Nanson, get third ranked, and so on.  This method
> finds a Condorcet order as well.  This is how I would use Nanson's
> method to get a full ranking.  I'll call it Nanson+.  If you invert
> the ballots, apply Nanson+, and invert the resulting ranking, you get
> inverse Nanson. 
> 
> I generally feel that it's up to a method's proponents to prove that a
> method has desirable properties, rather than up to others to prove
> that it doesn't.  Especially with a method that is rather difficult to
> carry out.
> 
> The inversing makes the method more confusing, and I doubt it improves
> on Nanson+'s properties.  In particular, I suspect the method violates
> monotonicity and clone independence.  It clearly violates reversal
> symmetry.  So, I think I'll stick with Ranked Pairs, at least until
> some desirable properties are proven for inverse Nanson.
> 
> ---
> Blake Cretney
> 
>