[EM] Friendly Voting: Some Criteria Compliance Proof Sketches

[Forest Simmons]

Suppose de granted each candidate one free bullet ballot .... that would
keep non-Smith candidates from being in the tied-for-win set ... but it
creates a non-scaling problem.

However, if we grant each candidate a gratuitous bullet ballot with
positive weight epsilon, and shrink epsilon until further shrinkage stops
changing the winner, then the method becomes scale invariant.

-Forest



On Sat, Oct 15, 2022, 2:18 PM Kristofer Munsterhjelm <km_elmet at t-online.de>
wrote:

> On 13.10.2022 02:38, Forest Simmons wrote:
> >
> >
> > On Wed, Oct 12, 2022, 3:35 AM Kristofer Munsterhjelm
> > <km_elmet at t-online.de <mailto:km_elmet at t-online.de>> wrote:
> >
> >     Does this come with the same caveat as in Friendly Cover that if
> >     someone
> >     has no first preference, then the compliance may be failed? E.g.
> >     suppose
> >     a bunch of nobodies are ranked first (enough so that they're not in
> >     Smith), then every viable candidate's first preference is zero.
> >
> >
> > Yeah, but if even one Smith member is in the support of lottery L, then
> > no non-Smith winner has a finite Sum S(X).
> >
> > That's why all my suggestions for Lare designed to take care of that
> > problem.
>
> My hunch is that since the strength of first preferences (against
> burial) comes from that X>W voters can't change who they vote first by
> burying, then a burial resistant extension should be based on first
> preferences of some subset of the candidates. (If we take the subset to
> be just every candidates, we get the usual Friendly Cover/Voting patterns.)
>
> If so, then always using the subset of every possible candidate will
> lead to the ISDA/pathological Smith/Landau failure shown above. So when
> comparing X to W, we have two possibilities: either a fixed-cardinality
> subset (probably containing both X and W) or a variable-cardinality one.
>
> IRV is "essentially" (if you handwave enough) doing the latter, with its
> path dependence producing nonmonotonicity and all of the usual flaws.
> Since variable-cardinality subsets seem to lead directly to summability
> violations, it's kind of a no-go. But perhaps there is a way to do the
> former and still retain burial resistance?
>
> There are two problems.
>
> First, suppose we always use subsets of three. Then Condorcet cycle
> analogs may occur - e.g. A's fpA-fpC score beats B's in the subset
> (A,B,C), B's beats C's in the subset (B, C, D), C's beats D's in the
> subset (C, D, A), and D's beats A's in the subset (D, A, B). The usual
> fix would be to use something like Ranked Pairs over these orderings,
> but I can't see how to prove that burial resistance will be preserved by
> this.
>
> Second, if we clone A into A1...An, then we can engineer the cloning to
> set the A clones' scores to arbitrary values for fpA-fpC restricted to
> the subset (A1,...,An). So using the result for subsets containing only
> clones to determine whether A or B should win would fail clone
> independence.
>
> So, not entirely easy. But perhaps it will give you some ideas for
> lotteries along the "restricted subset" path :-)
>
> Maybe the easiest way is to just come up with something that passes
> DMTCBR and is based on the fixed subset pattern, and see if its burial
> resistance is robust. A possible clue here is that fpA-fpC restricted to
> every (assume cardinality three) subset containing A will have A
> outscore B and C even after burial. But then again, that's also true of
> pairwise victories...
>
> Or it might be possible to do something along the lines of: if A ties B,
> then A's tiebreaker is based on the relation of A's max-scoring friend
> wrt B's -- because in the case that A covers B, then B can't possibly
> win. As long as the "relation of A's max-scoring friend wrt B's" also
> takes into consideration the tiebreakers of these max-scoring friends.
>
> > Also apologies for not checking with you before dubbing my version
> > "Friendly Voting."
> > I started out just trying to do generalized median voting, and was
> > surprised when the final simplified version turned out to be "Friendly"
> ; -)
>
> That's no trouble at all :-)
>
> I definitely won't complain if Friendly turns out to be a really good
> method!
>
> -km
>
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