[EM] Saari and Cyclic Ambiguities

[Arnold B. Urken]

Forest and Alex,

Just rejoined the list recently and enjoyed this message. I would like 
to point out that engineering and sports applications share a problem 
that was identified in Borda's lifetime: that voters truncate their 
rankings to avoid hurting their most favored choice. In baseball this seems to 
be accepted, but in selecting NASA mission trajectories, it took weeks to 
consult with the various teams to get a complete ordering.

In the Institute of France, Pierre Claude Daunou described this problem in some 
detail. The problem had serious practical consequences because the by-laws 
required waiting two weeks to take another vote--to avoid "cabals."  In any 
case, when Borda heard about this problem, he is reported to have said that his 
system was designed for "honest men."  

My apologies in advance if all of this is old news to you.

Arnie
Quoting Forest Simmons <fsimmons at pcc.edu>:

> Alex, I never got around to responding to this message of yours.  I just
> wanted to say that I thought your idea of cancelling all of the symmetries
> in the Smith set preference orders and then looking again for a CW  is a
> great idea!
> 
> It is one of those ideas like Martin Harper's Universal Approval and
> Richard Moore's Majority Potential that I wish I had thought of.
> 
> Forest
> 
> On Thu, 28 Mar 2002, [iso-8859-1] Alex Small wrote:
> 
> > I'll begin this message concerning Saari's insights by saying that, for
> the
> > record, I am AGAINST the Borda Count for real-world elections (although I
> > understand it has engineering and sports applications).  That said, let
> me
> > share a thought:
> >
> > Saari pointed out that cyclic ambiguities come from a "Condorcet profile"
> > or "symmetric profile".  If the electorate consists of 3 groups
> >
> > 35 A>B>C
> > 33 B>C>A
> > 32 C>A>B
> >
> > we can "decompose" the electorate into
> >
> > 32 A>B>C  +  3 A>B>C
> > 32 B>C>A  +  1 B>C>A
> > 32 C>A>B
> >
> > The total electorate has no Condorcet winner.  The "residual" does.
> >
> > It's my understanding that, in general, if there is no Condorcet winner,
> > subtracting out the symmetric profiles will leave a "residual" profile
> that
> > DOES have a Condorcet winner.
> >
> > Caveat:  If a Condorcet winner does exist, subtracting out the symmetric
> > profile might yield a different Condorcet winner.
> >
> > So, this leads me to propose the following possible election method:
> >
> > 1)  If a Condorcet winner exists then simply elect him.  End of story.
> > 2)  If not, eliminate from all ranking orders any candidate NOT in the
> > Smith set.  So, if A, B, and C are in the Smith set, a person whose
> > preference order is A>D>C>B is now considered to have the preference
> order
> > A>C>B.
> > 3)  Having narrowed down the list, subtract out the symmetric profiles.
> > 4)  There will now be a "Condorcet winner."  Elect him.
> >
> > A few weeks ago I discussed Saari's insight, and DEMOREP suggested that
> > discarding ballots like that is of dubious merit.  Here's my defense:
> >
> > Suppose that there's a conference on the mathematics of voting, and being
> > the election geniuses that we are we all give invited talks.  Ten of us
> > decide to go out to dinner together.  After we've delivered our talks,
> nine
> > people are waiting for me to show up (I'm running late, because I had to
> > call my fiance).
> >
> > Joe suggests that those present pick a restaurant.  The results are:
> >
> > 3 Thai>Pizza>Steak
> > 3 Pizza>Steak>Thai
> > 3 Steak>Thai>Pizza
> >
> > We have an exactly symmetric profile.  Demorep suggests approval voting,
> > but at that minute I walk through the door.  I maintain that _if_ we only
> > go by preference orders, then my vote should decide all.  Not because I'm
> > special, but because if you're only one vote away from an exact tie, that
> > single vote must be the deciding factor for any reasonable method.
> >
> > Since my preference order is Thai>Pizza>Steak I maintain that Thai should
> > be the winner.  With the method I proposed above Thai would win.  With
> SSD
> > the weakest defeat is Thai vs. Steak (margin of 2).  All other defeats
> have
> > a margin of 4.  SSD would hence select Thai.  Not knowing much about
> other
> > methods I can't say which others would pick Thai.
> >
> > So, it isn't such a stretch to conclude that cyclic ambiguities should be
> > resolved by whatever ballots are in excess of a tie.
> >
> > Of course, Demorep's idea of using approval to pick the winner
> (presumably
> > from among the members of the inner-most unbeaten set in the general
> case)
> > is far easier.  I merely throw this out for examination.  I haven't
> thought
> > out what flaws it might have.
> >
> > Alex
> >
> >
> 
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