[EM] A few more Bucklin variants, because why not?

[Etjon Basha]

Thanks Kristofer,

On why the CW must be in the Serious Candidate Set: once explained in
those terms, it really seems quite intuitive indeed, and I should have
thought a bit harder before posting.

I also think that explanation is sufficient to model the observed
relationship between the Smith and SC Sets: using your argument, setting
the approval cut-off at a Smith Set member will elect some Smith Set
member. Setting to the cutoff at some non-member may elect any candidate,
inside or outside the Smith Set.

These two suffice to explain what I've observed by random simulation: The
Smith and Serious Candidate Sets will share at least one member in common
and, beyond this, it seems like we may be able to say no further.

Regards,

Etjon Basha

On Fri, Sep 27, 2024 at 2:46 AM Kristofer Munsterhjelm <
km-elmet at munsterhjelm.no> wrote:

> On 2024-09-21 13:53, Etjon Basha wrote:
> > Dear gentlemen,
> >
> >
> > A while ago I did write here about the Iterated Bucklin
> > <https://electowiki.org/wiki/Iterated_Bucklin> method on which I’ve
> > recently had a chance to think and generalize about a bit more. Maybe
> > some of the below could be novel or otherwise of interest.
> >
> >
> > First, and for our purposes today, let's define the *Serious Candidates
> > Set* in the context of a ranked ballot, to include those candidates who
> > would win an approval count if they served as the approval cutoff across
> > all ballots.
> >
> >
> > In the [2:A>B, 3:C>A, 4:A>B] election as an example, the Set would
> > include A and B only, as applying the cutoff at C would still elect B.
> >
> >
> > I’ve been checking some random simulations from Kevin Venzke’s
> > votingmethods.net <http://votingmethods.net>, and here are some
> > properties of this Set that I *suspect*:
> >
> > 1.If there is a Condorcet Winner, this Set should always include them.
>
> I think that should hold, at least with strict ranks. Suppose A beats B
> pairwise. Then if we set the approval cutoff to A inclusive, every
> ballot that has B>A will give a point to B and A, whereas every ballot
> that has A>B will give a point only to A. Since A beats B pairwise, the
> number of A>B ballots exceed the number of B>A ballots. Thus A must have
> a higher approval score than B.
>
> Since A is the CW, this reasoning holds for all other B. A must obtain a
> higher approval score than B for any other B. Hence A is the winner and
> belongs to the SCS.
>
> > 2.Otherwise, this Set should always partially overlap with the Smith Set.
>
> The reasoning above gets us part of the way. Suppose A is in the Smith
> set. For any non-Smith member B, A must obtain more points than B for
> the same reason as above. So the non-Smith members can't win.
>
> What we have to show is that it's impossible for e.g. A to be the winner
> when the approval cutoff is set at B, B be the winner when it's set at
> C, and C be the winner when it's set at A. I'm not sure how to prove
> that, though.
>
> -km
>
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