[EM] Enhanced DMC

[fsimmons at pcc.edu]

Very good Chris.  

I tried to build a believable profile of ballots that would yield the approval order and defeats of this 
example without success, but I am sure that it is not impossible.

I think in general that if the approval scores are at all valid I would go for the enhanced DMC winner over 
any of the chain building methods we have considered.  I think other considerations over-ride the 
importance of being uncovered.

----- Original Message -----
From: "C.Benham" 
Date: Sunday, September 11, 2011 10:08 am
Subject: Enhanced DMC
To: election-methods-electorama.com at electorama.com
Cc: Forest W Simmons 

> Forest Simmons wrote (15 Aug 2011):
> 
> >Here's a possible scenario:
> >
> >Suppose that approval order is alphabetical from most approval 
> to least A, B, C, D.
> >
> >Suppose further that pairwise defeats are as follows:
> >
> >C>A>D>B>A together with B>C>D .
> >
> >Then the set P = {A, B} is the set of candidates neither of 
> which is pairwise
> >beaten by anybody with greater approval.
> >
> >Since the approval winner A is not covered by B, it is not 
> covered by any
> >member of P, so the enhanced version of DMC elects A.
> >
> >But A is covered by C so it cannot be elected by any of the 
> chain building
> >methods that elect only from the uncovered set.
> >
> 
> Forest,
> 
> Is the "Approval Chain-Building" method the same as simply 
> electing the 
> most approved uncovered candidate?
> 
> I surmise that the set of candidates not pairwise beaten by a 
> more 
> approved candidate (your set "P", what I've
> been referring to as the "Definite Majority set") and the 
> Uncovered set 
> don't necessarily overlap.
> 
> If forced to choose between electing from the Uncovered set and 
> electing 
> from the "DM" set, I tend towards
> the latter.
> 
> Since Smith//Approval always elects from the DM set, and your 
> suggested 
> "enhanced DMC" (elect the most
> approved member of the DM set that isn't covered by another 
> member) 
> doesn't necessarily elect from the Uncovered set;
> there doesn't seem to be any obvious philosophical case that 
> enhanced 
> DMC is better than Smith//Approval.
> 
> (Also I would say that an election where those two methods 
> produce 
> different winners would be fantastically unlikely.)
> 
> A lot of Condorcet methods are promoted as being able to give 
> the 
> winner just from the information contained in the
> gross pairwise matrix. I think that the same is true of these 
> methods 
> if we take a candidate X's highest gross pairwise
> score as X's approval score. Can you see any problem with that?
> 
> 
> Chris Benham
> 
> 
> 
> 
> ----- Original Message -----
> From:
> Date: Friday, August 12, 2011 3:12 pm
> Subject: Enhanced DMC
> To: election-methods at lists.electorama.com,
> 
> > > From: "C.Benham"
> > > To: election-methods-electorama.com at electorama.com
> > > Subject: [EM] Enhanced DMC
> >
> > > Forest,
> > > The "D" in DMC used to stand for *Definite*.
> >
> > Yeah, that's what we finally settled on.
> >
> > >
> > > I like (and I think I'm happy to endorse) this Condorcet method
> > > idea, and consider it to be clearly better than regular DMC
> > >
> > > Could this method give a different winner from the ("Approval
> > > Chain Building" ?) method you mentioned in the "C//A" thread 
> (on 11
> > > June 2011)?
> >
> > Yes, I'll give an example when I get more time. But for all 
> practical 
> > purposes they both pick the highest approval Smith candidate.
> 
> 
> 
> Here's a possible scenario:
> 
> Suppose that approval order is alphabetical from most approval 
> to least 
> A, B, C, D.
> 
> Suppose further that pairwise defeats are as follows:
> 
> C>A>D>B>A together with B>C>D .
> 
> Then the set P = {A, B} is the set of candidates neither of 
> which is 
> pairwise
> beaten by anybody with greater approval.
> 
> Since the approval winner A is not covered by B, it is not 
> covered by any
> member of P, so the enhanced version of DMC elects A.
> 
> But A is covered by C so it cannot be elected by any of the 
> chain building
> methods that elect only from the uncovered set.
> 
> 
> Forest Simmons wrote (12 June 2011):
> 
> > I think the following complete description is simpler than anything
> > possible for ranked pairs:
> >
> > 1. Next to each candidate name are the bubbles (4) (2) (1). The
> > voter rates a candidate on a scale from
> > zero to seven by darkening the bubbles of the digits that add 
> up to
> > the desired rating.
> >
> > 2. We say that candidate Y beats candidate Z pairwise iff Y 
> is rated
> > above Z on more ballots than not.
> >
> > 3. We say that candidate Y covers candidate X iff Y pairwise beats
> > every candidate that X pairwise
> > beats or ties.
> >
> > [Note that this definition implies that if Y covers X, then Y 
> beats X
> > pairwise, since X ties X pairwise.]
> >
> > Motivational comment: If a method winner X is covered, then the
> > supporters of the candidate Y that
> > covers X have a strong argument that Y should have won instead.
> >
> > Now that we have the basic concepts that we need, and 
> assuming that
> > the ballots have been marked
> > and collected, here's the method of picking the winner:
> >
> > 4. Initialize the variable X with (the name of) the 
> candidate that
> > has a positive rating on the greatest
> > number of ballots. Consider X to be the current champion.
> >
> > 5. While X is covered, of all the candidates that cover X, 
> choose the
> > one that has the greatest number of
> > positive ratings to become the new champion X.
> >
> > 6. Elect the final champion X.
> >
> > 7. If in step 4 or 5 two candidates are tied for the number of
> > positive ratings, give preference (among the
> > tied) to the one that has the greatest number of ratings 
> above level
> > one. If still tied, give preference
> > (among the tied) to the one with the greatest number of 
> ratings above
> > the level two. Etc.
> >
> > Can anybody do a simpler description of any other Clone Independent
> > Condorcet method?
> 
> 
> 
>