Finding the probable best candidate?

Steve Barney barnes992001 at yahoo.com
Fri Feb 22 17:13:56 PST 2002


Forest:

Here is a quote from one of Saari's papers which seems to imply that inverse
Nanson and the Borda Bubble Sort method must not be monotonic, given that they
are based on an iterative Borda Count applied to subsets of candidates:

"To illustrate, the development of Chapter 5 from [_The Basic Geometry of
Voting_] proves that Nanson’s method is not monotonic. If, however, only Basic
profiles are used, the Nanson’s method is monotonic. This lack of monotonicity
is caused by the Condorcet portion of a profile. . . . The BC applied to the
set of all n candidates is susceptible only to Basic profiles. For subsets of
candidates, the BC is susceptible to Basic and Condorcet profiles..."
--Pg 32 in Donald Saari, "Mathematical structure of voting paradoxes? II.
Positional voting," _Economic Theory_, Vol. 15, pgs 55–102 (January, 2000).

So, this may not be the case, but I guess that some, if not all, profiles which
produce a violation of monotonicity with Nanson's method, will also produce a
violation of monotonicity with other iterative Borda Count methods. Can you
prove otherwise by showing us an example where Nanson's violates monotonicity,
but Inverse Nanson or Bubble Sort Borda do not?

SB

--- In election-methods-list at y..., Forest Simmons <fsimmons at p...> wrote:
> On Wed, 20 Feb 2002, Rob LeGrand wrote:
> 
> > Forest wrote:
[...]
> > What is Inverse Nanson again?  How is it different from regular Nanson (or
> > Baldwin)?  Is it monotonic?
> > 
> 
> Sets aside the Borda winner, then sets aside the Borda winner of the
> remaining, etc. down to until one candidate is left.  That candidate is
> the first to be eliminated.
> 
> This first eliminated candidate will always be in the "reverse smith set."
> 
> The process is repeated among the uneliminated candidates, until there is
> only one uneliminated candidate, the Inverse Nanson winner, which will
> always be a member of the smith set.
> 
> I think it is monotone, but I haven't proven it.
> 
> Forest


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