[EM] Improved Copeland

Ted Stern dodecatheon at gmail.com
Wed Jun 12 11:14:56 PDT 2019

```Hi Forest,

Try this example:

98: Abby >  Cora >  Erin >  Dave > Brad
64: Brad >  Abby >  Erin >  Cora > Dave
12: Brad >  Abby >  Erin >  Dave > Cora
98: Brad >  Erin >  Abby >  Cora > Dave
13: Brad >  Erin >  Abby >  Dave > Cora
125: Brad >  Erin >  Dave >  Abby > Cora
124: Cora >  Abby >  Erin >  Dave > Brad
76: Cora >  Erin >  Abby >  Dave > Brad
21: Dave >  Abby >  Brad >  Erin > Cora
30: Dave >  Brad >  Abby >  Erin > Cora
98: Dave >  Brad >  Erin >  Cora > Abby
139: Dave >  Cora >  Abby >  Brad > Erin
23: Dave >  Cora >  Brad >  Abby > Erin

except Dave.  So S(Abby) is the total number of ballots on which Brad is
ranked above all other candidates except possibly Dave.  So T(Brad) = Brad
> Abby votes , 463 minus the 23 ballot where Cora > Brad.  So S(Abby) = 440.

Similarly, S(Brad) = T(Dave).  Dave defeats all candidates except Abby and
Erin.  So T(Dave) = total number of ballots on which Dave is ranked higher
than all candidates except possibly Abby or Erin.  So T(Dave) = 21 + 30 +
98 + 139 + 23 = 311.

I haven't worked out the rest, but I believe they are all higher.  So Brad
would beat Abby.  I think in this case I would prefer Abby to Brad, so I'm
not entirely happy.

I am also not seeing an obvious way to make this summable.

What's wrong with using Equal-Rated-Zero for pairwise votes, and then
minimizing the sum of defeating scores against a candidate?  In other
words, add up all the defeating scores in a candidate's column in the
pairwise array.  I'm sure that has a name already. In this example,  Abby
would win with a total of 463 votes Brad>Abby, which is less than the total
defeating scores for any other candidate.  It also has the advantage of
being summable with no other information than the pairwise array required.

On Tue, Jun 11, 2019 at 5:02 PM Forest Simmons <fsimmons at pcc.edu> wrote:

> Our first attempts at improved Copeland ended up losing monotonicity
> without fully achieving clone independence.
>
> I will repeat that version here for comparison:
>
> Elect the candidate with the fewest top rank ballot totals for the
> candidates that beat her pairwise.
>
> It turns out that we have to replace the top rank totals with something
>
> For each candidate X let T(X) be the number of ballots on which candidate
> X is ranked above all of the candidates that she beats pairwise.
>
> This total includes all of the unique top votes of candidate X, but also
> includes some others.
>
> So here's the method: elect the candidate Y that minimizes S(Y) defined as
> the sum of T(X) (over all X that beat Y pairiwise).
>
> Here's an example:
>
> 4 A>B
> 2 B>C
> 3 C>A
>
> T totals in the form of top votes plus extras from second ranks:
> T(A) = 4 + 3 = 7
> T(B) = 2 + 4 = 6
> T(C) = 3 +2 = 5
>
> S(A) = T(C) = 5 < 6 = T(B)=S(C) < 7 = T(A)=S(B),
>
> so S(A) < S(C) < S(B)
>
> A is the winner, B is the loser, and C is in the middle of the social
> order according to this method.
>
> I would appreciate it being tested on your favorite examples. If it needs
> clarification I will use your examples to clarify it, as long as it holds
> up to scrutiny.
>
> Thanks,
>
> Forest
>
>
>
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>
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