[EM] Decloned Copeland Borda Hybrid

[Forest Simmons]

 James,

 you are absolutely right ... as I stated it the method is basically just
Implicit Approval ... definitely not Condorcet efficient.

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.

Thanks for taking the time to check this out!

Forest

El mié., 6 de abr. de 2022 6:55 a. m., James Faran <jjfaran at buffalo.edu>
escribió:

> I may not understand the method, but consider the following with
> candidates X, Y, Z:
>
> 51 X<Y<Z
> 49 Y>Z>X
>
> Then X is the CW, and X is ranked over or pairwise defeats a'(B) only on
> those ballots B in the group of 51. Thus N(X)=51.
>
> Meanwhile, Y is ranked over a'(B) on all ballots, so N(Y)=100.
>
> If I've understood the method  correctly, it is not Condorcet efficient.
>
> Jim Faran
> ------------------------------
> *From:* Election-Methods <election-methods-bounces at lists.electorama.com>
> on behalf of Forest Simmons <forest.simmons21 at gmail.com>
> *Sent:* Wednesday, April 6, 2022 2:57 AM
> *To:* EM <Election-methods at lists.electorama.com>; Kristofer Munsterhjelm <
> km_elmet at t-online.de>
> *Subject:* [EM] Decloned Copeland Borda Hybrid
>
> 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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