[EM] Saari and Cyclic Ambiguities

[Forest Simmons]

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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