[Election-Methods] Using range ballots as an extension of ranked ballot voting
Michael Rouse
mrouse1 at mrouse.com
Sat Mar 15 10:55:50 PDT 2008
Let me know if this is formatted incorrectly -- it word wraps just fine
in Thunderbird, but I don't know if other email programs display it
correctly.
Sorry it's taken so long to reply -- it's been a hectic week.
I also apologize for the length of the message, which probably causes
eyes to glaze over and the message quickly deleted (grin). I marked the
results of each calculation and things I considered more interesting
with four asterisks (****), so people can skip over the rather tedious math.
Anyway...
Juho wrote:
> 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.
On the other hand, it wouldn't make either A1 or A2 the winner
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
A1=A2 (tie)
A1>B (1)
A2>B (1)
A1>C (1)
A2>C (1)
B>C (1)
C>D (3)
D>A1 (3)
D>A2 (3)
D>B (3)
To make A1 the weak Condorcet winner (A1>=A2,B,C,D), removing the
relation D>A is sufficient. The total distance is 12.
To make A2 the weak Condorcet winner (A2>=A1,B,C,D), removing the
relation D>A is sufficient. The total distance is 12.
To make B the weak Condorcet winner (B>=A1,A2,C,D), removing the
relation A1>B, A2>B, and D>B is sufficient. The total distance is 17
(1+1+15)
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)
To make D the weak Condorcet winner (D>=A1,A2,B,C), removing the
relation C>D is sufficient. The total distance is 11.
****In this case, cloning A would give D the victory.
Let's check the other cloning possibilities (I make no guarantees on my
math -- I cut and pasted a bit to save on typing, and I might have
messed up. Still, it *looks* okay to me):
If we clone B instead:
1: A=10 B1=2 B2=2 C=1 D=0
1: A=10 C=7 B1=6 B2=6 D=0
1: B1=10 B2=10 C=6 A=5 D=0
3: C=10 D=5 A=1 B1=0 B2=0
3: D=10 B1=4 B2=4 A=3 C=0
B1=B2
A>B1 (1)
A>B2 (1)
A>C (1)
B1>C (1)
B2>C (1)
C>D (3)
D>A (3)
D>B1 (3)
D>B2 (3)
To make A the weak Condorcet winner (A>=B1,B2,C,D), removing the
relation D>A is sufficient. The total distance is 12.
To make B1 the weak Condorcet winner (B1>=A,B2,C,D), removing the
relation A>B1 and D>B1 is sufficient. The total distance is 16
To make B2 the weak Condorcet winner (B2>=A,B1,C,D), removing the
relation A>B2 and D>B2 is sufficient. The total distance is 16
To make C the weak Condorcet winner (C>=A,B1,B2,D), removing the
relation A>C, B1>C, and B2>C is sufficient. The total distance is 5
To make D the weak Condorcet winner (D>=A,B1,B2,C), removing the
relation C>D is sufficient. The total distance is 11
****C is still the winner.
And cloning C:
1: A=10 B=2 C1=1 C2=1 D=0
1: A=10 C1=7 C2=7 B=6 D=0
1: B=10 C1=6 C2=6 A=5 D=0
3: C1=10 C2=10 D=5 A=1 B=0
3: D=10 B=4 A=3 C1=0 C2=0
C1=C2
A>B (1)
A>C1 (1)
A>C2 (1)
B<C1 (1)
B<C2 (1)
C1>D (3)
C2>D (3)
D>A (3)
D>B (3)
To make A the weak Condorcet winner (A>=B,C1, C2,D), removing the
relation D>A is sufficient. The total distance is 12.
To make B the weak Condorcet winner (B>=A,B,C1,C2,D), removing the
relation A>B and D>B is sufficient. The total distance is 16.
To make C1 the weak Condorcet winner (C1>=A,B,C2,D), removing the
relation A>C1 and B>C1 is sufficient. The total distance is 4.
To make C2 the weak Condorcet winner (C2>=A,B,C1,D), removing the
relation A>C2 and B>C2 is sufficient. The total distance is 4.
To make D the weak Condorcet winner (D>=A,B1,B2,C), removing the
relation C1>D and C2>D is sufficient. The total distance is 22.
****This drops D way down, but the clone C1 and C2 are still the winners.
And finally, cloning D:
1: A=10 B=2 C=1 D1=0 D2=0
1: A=10 C=7 B=6 D1=0 D2=0
1: B=10 C=6 A=5 D1=0 D2=0
3: C=10 D1=5 D2=5 A=1 B=0
3: D1=10 D2=10 B=4 A=3 C=0
D1=D2
A>B (1)
A>C (1)
B>C (1)
C>D1 (3)
C>D2 (3)
D1>A (3)
D2>A (3)
D1>B (3)
D2>B (3)
To make A the weak Condorcet winner (A>=B,C,D), removing the relation
D1>A and D2>A is sufficient. The total distance is 24.
To make B the weak Condorcet winner (B>=A,C,D), removing the relation
A>B is sufficient. The total distance is 16.
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.
To make D1 the weak Condorcet winner (D1>=A,B,C,D2), removing the
relation C>D1 is sufficient. The total distance is 11.
To make D2 the weak Condorcet winner (D2>=A,B,C,D1), removing the
relation C>D2 is sufficient. The total distance is 11.
****C is the winner in this one as well.
>
> 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
>
>
>
I hesitate to add extra weight to the "worst defeat" (though I may be
misunderstanding what you mean by that), because I don't want to give an
incentive to someone trying to bury a candidate.
In fact, to help to prevent burying, I was looking at a couple of
different methods -- dropping the lowest-ranked but equivalent relations
(A=5, C=0 would be dropped before B=10, A=5, even though they are the
same distance apart), and dropping the lowest ranked no matter what
distance apart (dropping C=7, D=0 before A=10, C=7). That should remove
some of the incentive to bury a candidate.
I'm also playing with one where you take the median of all distances in
a relation, and compare pairwise from that. As always, no claims for
originality are made (I don't remember it offhand, but I may have
forgotten). Note that what follows is a bit of random musing that
occurred to me while writing, so I thought I'd check it out.
Using the original 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
Here are the distances for each pair relation:
A>B (8,4,-5,1,1,1,-1,-1,-1)
Ordered highest-lowest: 8,4,1,1,1,-1,-1,-1,-5. Median is 1.
A>C (9,3,-1,-9,-9,-9,3,3,3)
Ordered highest-lowest: 9,3,3,3,3,-1,-9,-9,-9. Median is 3.
A>D (10,10,5, -4,-4,-4,-7,-7,-7)
Ordered highest-lowest: 10,10,5,-4,-4,-4,-7,-7,-7. Median is -4.
B>A (-8,-4,5,-1,-1,-1,1,1,1)
Ordered highest-lowest: 5,1,1,1,-1,-1,-1,-4,-8). Median is -1.
B>C (1,-1,4,-10,-10,-10,4,4,4)
Ordered highest-lowest: 4,4,4,4,1,-1,-10,-10,-10. Median is 1.
B>D (2,6,10,-5,-5,-5,-6,-6,-6)
Ordered highest-lowest: 10,6,2,-5,-5,-5,-6,-6,-6. Median is -5.
C>A (-9,-3,1,9,9,9,-3,-3,-3)
Ordered highest-lowest: 9,9,9,1,-3,-3,-3,-3,-9. Median is -3.
C>B (-1,1,-4,10,10,10,-4,-4,-4)
Ordered highest-lowest: 10,10,10,1,-1,-4,-4,-4,-4. Median is -1.
C>D (1,7,6,5,5,5,-10,-10,-10)
Ordered highest-lowest: 7,6,5,5,5,1,-10,-10,-10. Median is 5.
D>A (-10,-10,-5,4,4,4,7,7,7)
Ordered highest-lowest: 7,7,7,4,4,4,-5,-10,-10. Median is 4.
D>B (-2,-6,-10,5,5,5,6,6,6)
Ordered highest-lowest: 6,6,6,5,5,5,-2,-6,-10. Median is 5.
D>C (-1,-7,-6,-5,-5,-5,10,10,10)
Ordered highest-lowest: 10,10,10,-1,-5,-5,-5,-6,-7. Median is -5.
The positive median distances ordered from highest to lowest are:
C>D, D>B (5)
D>A (4)
A>C (3)
A>B, B>C (1)
If there is a Condorcet winner, it will have a positive median distance
in each pair. I haven't quite decided how to break cycles in this
method, though you can look at the various possible orders and see which
relation has the highest score (positive if it supports the relation,
negative if it does not), e.g.:
A>B>C>D 1+3-4+1-5+5 = 1
A>B>D>C 1-4+3-5+1-5 = -9
A>C>B>D 3+1-4-1+5-5 = -1
A>C>D>B 3-4+1+5-1+5 = 9
A>D>B>C -4+1+3+5-5+1 = 1
A>D>C>B -4+3+1-5+5-1 = -1
B>A>C>D -1+1-5+3-4+5 = -1
B>A>D>C -1-5+1-4+3-5 = -11
B>C>A>D 1-1-5-3+5-4 = -7
B>C>D>A 1-5-1+5-3+4 = 1
B>D>A>C -5 -1+1+4-5+3 = -3
B>D>C>A -5+1-1-5+4-3 = -9
C>A>B>D -3-1+5+1-4-5 = -7
C>A>D>B -3 +5-1-4+1+5 = 3
C>B>A>D -1-3+5-1-5-4 = -9
C>B>D>A -1+5-3-5-1+4 = -1
C>D>A>B 5-3-1+4+5-1 = 9
C>D>B>A 5-1-3+5+4-1 = 9
D>A>B>C 4+5-5+1+3+1 = 9
D>A>C>B 4-5+5+3+1-1 = 7
D>B>A>C 5+4-5-1+1+3 = 7
D>B>C>A 5-5+4+1-1-3 = 1
D>C>A>B -5+4+5-3-1+1 = 1
D>C>B>A -5+5+4-1-3+4 = 4
Unfortunately, the highest score (9) is not unique. You have:
A<C<D<B
C>D>A<B
C>D>B>A
D>A>B>C
Interestingly, the unique "biggest loser" is B>A>D>C, the inverse of
C>D>A>B (one of the top 4 winners). It might be fun to check the
greatest difference between winners and their inverse, and come up with
an order from the comparison.
What I think is especially interesting about medians is that it seems
fairly stable even with wide swings in an individual ballot -- if you
score someone less than the median, it doesn't matter if it's a tiny bit
or if you tried to bury it. The only time your individual vote will
change the median is if you cross it, and it will affect it the same way
-- if you change your vote to the opposite side of the median, the
median can either stay the same or move the same direction, but it will
never go the opposite direction.
I apologize once again for he length of this post. If you come up with
an interesting variation or something brand new, let me know!
Michael Rouse
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