[EM] Reverse Symmetry Criterion

[Forest Simmons]

Martin,

Actually, I didn't quite say what I meant in my previous reply. What I
meant to say is that there should be no stigma in failing the RSC when the
entire set of candidates form a Condorcet cycle.

It would be an interesting senior level linear algebra project to see if
the following is true:

Conjecture:  If all of the candidates together form a Condorcet cycle,
then it is possible to replace all of the preference ballots with
consistent CR ballots such that when all of the CR ballots are averaged
together all of the candidates end up with the same average rating.

Forest

On Tue, 27 Mar 2001, Martin Harper wrote:

> Forest Simmons wrote:
> 
> > I like your idea.  But I still think there is no stigma attached to
> > failing the reverse symmetry criterion when there is no (unique) Condorcet
> > winner, so I would suggest that you throw those out of the count.
> >
> > Forest
> 
> Hmm - I'm not entirely convinced. It's certainly more reasonable to fail RSC when
> there's no condorcet winner - but I'd rather state this the other way: it's hugely
> bad to fail RSC when there's a condorcet winner, and methods which do so should be
> avoided like the plague. However, all other things being equal, I'd say that a
> method which passes RSC is better than one that passes RSC only when there's a
> condorcet winner.
> 
> Here's a stupid example:
> 
> 11 A>B>C>D>E>F
> 10 B>C>A>E>F>D
> 9 C>A>B>F>D>E
> 
> Now there's neither a Condorcet Winner, nor a Condorcet Loser, but I reckon any
> method which elects the same person as both the best and worst candidate has to
> have made a mistake somewhere...
> 
> On a side note, it's possible that a set of votes has a CW, but no CL (or vica
> versa). Would you judge such situations as having no stigma either? Or to have less
> stigma than the case where there is both a unique CW and CL?
> 
>