[EM] Decloned Copeland Borda Hybrid

[Forest Simmons]

Kevin,

I'm glad you caught that. It looks like we have to go back to explícit
designations of anti-favorites to avoid this "accidental" lowering of B on
a set of ballots resulting from intentional raising of C on other ballots.

The other option would be to count truncated votes fractionally in the
anti-favorite tallies. But that would be messy in hand computations.

The explicit option seems better at distinguishing "don't care" truncations
from strongly felt anti-favorite truncations ... for, e.g. avoiding the
"dark horse" back door that low VSE candidates might otherwise sneak
through.

Also, instead of adding implicit approval points when X is raised from
bottom, it turns out to be smoother to subtract points from anti-favorites
(as we shall see below).

So this method becomes ...

Elect the candidate with the greatest difference between the number of
ballots on which the designated anti-favorite is among those defeated
pairwise by it and the number of ballots on which it (itself) is the
designated anti-favorite.

The dual hybrid method is ...

Elect the candidate with the greatest difference between the number of
ballots on which it (itself) is the designated favorite and the number of
ballots on which it is pairwise defeated by the designated favorite.

Each of these mutually dual methods is a stand alone monotonic clone-free
method with high (but not perfect) Condorcet efficiency.

They can be combined into a full-blown high-falutin hybrid method where the
Borda component is more apparent:

Elect the candidate that maximizes the number of ballots on which it is the
designated favorite Plus the number of ballots on which it pairwise defeats
the designated anti-favorite Minus the number of ballots on which it is the
designated anti-favorite Minus the number of ballots on which it is
pairwise defeated by the designated favorite.

Note that with complete rankings in the three candidate case the first and
third terms of this expression together amount to the Borda Count of the
candidate in question.

It turns out that in such a three candidate case when there are only three
ballot factions and no Condorcet candidate, the Borda Count decides the
finish order.

For example ...

45 A>B>C
35 B> C>A
20 C>A>B

[No need to explicify favorites or anti-favorites where there is no
ambiguity]

The respective candidate scores are the Borda Counts

45-35, 35-20, and 20-45,

while all of the other terms precisely cancel each other.

The non-perfect Condorcet efficiency may well be a blessing in disguise
because it answers a common complaint that "... Condorcet methods have no
mechanism for weeding out low approval/VSE CW's."

Alternatively, modify the method to read ... "if there is no pairwise
undefeated candidate, then elect the candidate that maximizes ...."

-Forest



El jue., 7 de abr. de 2022 11:47 a. m., Kevin Venzke <stepjak at yahoo.fr>
escribió:

> Hi Forest,
>
> I'm not sure if you're still pursuing this method but as I understand it I
> don't
> think it's monotone.
>
> 0.394: B>C>A
> 0.366: A
> 0.151: C
> 0.087: A>B>C
>
> A>B>C>A cycle. Effective last preferences are A, B, A, C. Then C wins.
>
> Switch last faction to
> 0.087: A>C>B
>
> Cycle remains. Effective last preferences are now A, B, B, B. This causes
> C to net
> lose a few points, and A to gain a lot due to the increased relevance of
> the A>B
> contest, so that A wins.
>
> The mechanism also doesn't seem to ensure majority favorite, but maybe you
> know that.
>
> I do think the "anti-favorite" concept is interesting.
>
> Kevin
>
>
> Le mercredi 6 avril 2022, 14:53:01 UTC−5, Forest Simmons <
> forest.simmons21 at gmail.com> a écrit :
> > I intended to say each candidate ranked above or pairwise beating a'
> gets a point, and two points for both.
> >
> > The monotonicity proof goes through unchanged.... as z goes down, y
> gains a point, but so does X as it goes up.
>
> > > On each ballot B identify the anti-favorite a'(B) as the candidate a'
> among those bottom listed on ballot B that is bottom listed the most on the
> rest of the ballots.
> > >
> > > [A candidate is bottom listed on a ballot if it is ranked over no
> candidate on that ballot]
> > >
> > > Elect the candidate X that, on the most ballots B, either is ranked
> ahead of or pairwise defeats a'(B).
> > >
> > > In other words, elect argmax N(X), where N(X) is the cardinality of
> the set of ballots
> > >
> > > {B | X is ranked ahead of or pairwise defeats a'(B) (or both)}.
> > >
> > > The key to monotonicity is that if some candidate X is uniquely raised
> on some ballot B, then for any other candidate Y, the difference N(X)-N(Y)
> does not decrease.
> > >
> > > The only non-trivial case is where X starts in the a'(B) position and
> swaps positions with some Z that was ranked above it.
> > >
> > > This definitely adds a point to N(X) because X is newly ranked ahead
> of a'(B). This will also add a point to Y if Z is pairwise defeated by Y.
> > >
> > > In any case, N(X)-N(Y) does not decrease.
> > >
> > >  Does this method satisfy the Condorcet Criterion?
> > >
> > > Well, if X pairwise defeats all other candidates, then N(X) is equal
> to the total number of ballots minus the number of ballots on which X is in
> the a' position, which must be fewer than half of the ballots, otherwise X
> would be the Condorcet loser.
> > >
> > > Is it possible for N(Y) to be greater than this N(X) while losing
> pairwise to the CW  X?
> > >
> > > Perhaps ... if so, then this method is not a stand alone Condorcet
> method.
> > >
> > > It would be a valuable task for someone to test its Condorcet
> efficiency experimentally.
> > >
> > > Thanks!
> > >
> > > -Forest
>
>
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