[Election-Methods] Using range ballots as an extension of ranked ballot voting

[Juho]

The explanation was clear enough for me.

I have one concern - the behaviour of the counting method with  
clones. Let's multiply one of the candidates (A => A1 and A2). Then  
we would have:
1: A1=10 A2=10 B=2 C=1 D=0
1: A1=10 A2=10 C=7 B=6 D=0
1: B=10 C=6 A1=5 A2=5 D=0
3: C=10 D=5 A1=1 A2=1 B=0
3: D=10 B=4 A1=3 A2=3 C=0

Before the modification the results of C were:
> To make C the weak Condorcet winner (C>=A,B,D), removing the  
> relation A>C and B>C is sufficient. The total distance is 4 (3+1)


Is it so that with the new votes the results of C would be:
To make C the weak Condorcet winner (C>=A1,A2,B,D), removing the  
relation A1>C, A2>C and B>C is sufficient. The total distance is 7 (3 
+3+1)

This would mean that by naming numerous candidates one party (A1,A2)  
could make the position of some other party (C) worse.

One alternative approach would be to measure only the worst defeat,  
not the sum of all defeats.


Another (more philosophical) concern is that when counting the  
results of C (with the original votes) the distance is counted so  
that one can say that by electing C we violate the interests of voter  
"A=10 B=2 C=1 D=0" (=partly justified claim that "B would have won C  
in a pairwise comparison") by 1 point (if we look at how the method  
calculated the sum). This voter may however not be satisfied with  
this explanation since he/she may now say that C lost also to A and  
actually his/her interests (=partly justified claim that "A would  
have won C") were actually violated by 9 points.

My point is that if the intention is to justify the method by the  
small amount of damage the selections cause to the voters, there are  
different ways to measure the level of damage. Now the method makes  
the comparisons based on individual pairwise comparisons in  
individual votes, but it could also look at the damage e.g. at a per  
voter level.

Juho


On Mar 9, 2008, at 3:40 , <mrouse1 at mrouse.com> <mrouse1 at mrouse.com>  
wrote:

> Snipping the message:
>
>
> **********************
>
> On Mar 3, 2008, at 1:45 , <mrouse1 at mrouse.com> <mrouse1 at mrouse.com>  
> wrote:
>
>
> > juho4880 at yahoo.co.uk:
> >
>
> > >>Can you also clarify a bit how step 3 is counted when some  
> candidate X is beaten by two other candidates (Y and Z).
> > >>I find the proposed method interesting since it seems to aim at  
> electing good winners (using a function minimizes the problems  
> caused to the voters, from one point of view).
> >
> > I'd be happy to try. Do you have an example election for me to  
> play with? I'm assuming you mean where I said
> >
> >
> > 3. If there is no Condorcet winner, find the shortest distance  
> (sum of individual ranges) necessary to produce a Condorcet winner.
> >
> Sorry for some delay in replying. Here's one quick example.
>
> 1: A=10 B=2 C=1 D=0
> 1: A=10 C=7 B=6 D=0
> 1: B=10 C=6 A=5 D=0
> 3: C=10 D=5 A=1 B=0
> 3: D=10 B=4 A=3 C=0
>
> C is now beaten by both A and B, and C has to win them both in  
> order to become a Condorcet winner. What is the "shortest distance  
> (sum of individual ranges)" for C in this example and how do you  
> count it?
> **************
>
>
> Okay, here's how I did it by hand (sorry if it's a bit cryptic).
>
>
> Given the following:
>
> 1: A=10 B=2 C=1 D=0
> 1: A=10 C=7 B=6 D=0
> 1: B=10 C=6 A=5 D=0
> 3: C=10 D=5 A=1 B=0
> 3: D=10 B=4 A=3 C=0
>
> The question was: "C is now beaten by both A and B, and C has to  
> win them both in order to become a Condorcet winner. What is the  
> "shortest distance (sum of individual ranges)" for C in this  
> example and how do you count it?"
>
>
> Here are the number of pairs each way:
> A>B (1+1+3)
> A>C (1+1+3)
> A>D (1+1+1)
> B>A (1+3)
> B>C (1+1+3)
> B>D (1+1+1)
> C>A (1+3)
> C>B (1+3)
> C>D (1+1+1+3)
> D>A (3+3)
> D>B (3+3)
> D>C (3)
>
> Simplifying (numbers in parenthesis indicate surplus votes) and  
> showing the pair relations:
> A>B (1)
> A>C (1)
> B<C (1)
> C>D (3)
> D>A (3)
> D>B (3)
>
> To remove the relation A>B, it would take 1 vote, the smallest  
> total distance of which is 1 (1-0).
> To remove the relation A>C, it would take 1 vote, the smallest  
> total distance of which is 3 [(10-7) or (3-0)]
> To remove the relation B>C, it would take 1 vote, the smallest  
> total distance of which is 1 (2-1)
> To remove the relation C>D, it would take 3 votes, the smallest  
> total distance of which is 11 [(1-0)+(10-5)+(10-5)]
> To remove the relation D>A, it would take 3 votes, the smallest  
> total distance of which is 12 [(5-1)+(5-1)+(5-1)]
> To remove the relation D>B, it would take 3 votes, the smallest  
> total distance of which is 15[(5-0)+(5-0)+(5-0)]
>
> To make A the weak Condorcet winner (A>=B,C,D), removing the  
> relation D>A is sufficient. The total distance is 12.
> To make B the weak Condorcet winner (B>=A,C,D), removing the  
> relation A>B and D>B is sufficient. The total distance is 16 (1+15)
> To make C the weak Condorcet winner (C>=A,B,D), removing the  
> relation A>C and B>C is sufficient. The total distance is 4 (3+1)
> To make D the weak Condorcet winner (D>=A,B,C), removing the  
> relation C>D is sufficient. The total distance is 11.
>
> Using this method, C would be the winner, since 4 is the shortest  
> distance. The complete order is C>D>A>B.
>
> (I use the weak Condorcet criterion, because an infinitesimal  
> amount added to either candidate in a tie is sufficient to create a  
> winner.)
>
>
> Let me know if anything is unclear, and I'll try to give a better  
> explanation (grin).
>
>
> I might play around with the same election and see what removing  
> the lowest order of preferences (and not just the closest  
> preferences) would yield.
>
> Michael Rouse.
>
>
> ----
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