[EM] 3 or more choices - Condorcet

[Kristofer Munsterhjelm]

On 10/01/2012 12:13 AM, Juho Laatu wrote:
> On 30.9.2012, at 15.41, Kristofer Munsterhjelm wrote:

>> As far as intrinsically Condorcet methods go, Ranked Pairs feels
>> simple to me. The only tricky part is the indirect nature of the
>> "unless it contradicts what you already affirmed" step.
>
> To me the biggest problem of path based methods is that there is no very
> good real life explanation to why chains of pairwise victories are so
> important. In real life the idea of not electiong a candidate that would
> lose to someone who would lose to someone etc. doesn't sound like an
> important criterion (since it doesn't talk about what the candidate is
> like or how strong the opposition would be, but about what the set of
> candidates and its network of relations looks like). Probably there will
> never be a long chain of changes from one winner to another in real life.

I don't think you need to go into path logic for Ranked Pairs. Rather, 
how about this?

"Because of the existence of cycles, it's obvious we need to discard 
some of the data. So, what data do we discard? If we have to discard a 
one-on-one victory, lets discard those that are as narrow, or involve as 
few voters, as possible. Hence, we should go down the list of one-on-one 
contests and add the data they give to our order unless it would produce 
a cycle. That way, all the decisive contests get counted first and if we 
have to throw some away, it's the weaker ones."

It's a little IRVish (justifying the method by the way it works rather 
than the outcome), but still...