[EM] Margins Sorted Top-Ratings

[Ted Stern]

On Mon, Jan 29, 2024 at 10:55 AM C.Benham <cbenham at adam.com.au> wrote:

> (This is a re-send with an error corrected)
>
> I think  Margins Sorted Top Ratings would be a good  relatively burial
> resistant public proposal.
>
> * Voters rank from the top however many candidates they wish.
> Equal-ranking allowed.
>
> Give each candidate a score according to the number of ballots on which
> they are ranked below
> no others.
>
> Line them up in that order, highest to lowest.
>
> Check to see if all the candidates above bottom in this order pairwise
> beat the candidate immediately
> below them.
>
> If they do then elect the candidate highest in the order.
>
> If not begin with the pair that is pairwise out of order by the lowest
> margin and swap them.
> (if there is an exact tie in the size of the margin then swap the
> tied-margin pair lowest in the order).
>
> Repeat until no pair of adjacent candidates are pairwise out of order
> and then elect the highest-ordered
> candidate. *
>
> This could also use ratings ballots.
>
> This meets Condorcet, but can be at least be explained (if not sold)
> without reference to Condorcet or Smith.
>
> It would be as monotonic as it is possible for a Condorcet method to be.
>
> For the sake of simplicity (and elegance) it has some short-comings.
> When there is a top cycle, voters who
> didn't top-rate (rank below no other candidates) any of the candidates
> in the Smith set are disadvantaged by comparison
> those that did.  It would also fail Clone-Independence.
>

Could you explain how Top-rated Sorted Margins fails Independence from
Clones? I'm not seeing how that would happen, unless you have unusual
restrictions on top-rating


>
> A much more complicated method idea I had (that would be the same thing
> with three candidates):
>
> *Voters rank from the top however many candidates they wish.
> Equal-ranking allowed.
>
> (1) Eliminate (drop from the ballots and henceforth ignore) all
> candidates not in the Smith set.
>
> (2) Score the remaining candidates according to their minimum pairwise
> scores, with ballots that rank two candidates
> equal-top contributing a whole vote to each of the two candidate's
> scores against each other. Otherwise ballots that
> rank two candidates equal below top contribute zero to their pairwise
> scores against each other.
>
> (A possible variation is that they contribute half a vote to each if
> they are ranked below top and above bottom.)
>

This idea is interesting, but I'd like to see a more nuanced incorporation
of the strength of equal ratings.

The simplest way to do that would be Equal-Rated-Score-fraction. In other
words, when two candidates are equally rated, they both receive
score/maxscore of a vote. So an equal top pair would each receive 1, and
equal bottom pair would receive zero, and equal-in-between would receive
something in between.

And of course, the scores would have to be recalculated after the Smith
elimination step. Say you had Smith candidates A3=B3 > C1=D1 on a 0-5
rating, and A, B, C were in Smith. Then you'd shift to A5=B5 > C3=D3, and
score A>B:1, B>A:1, C>D:0.6, D>C:0.6.


>
> (3) Eliminate all candidates that don't have a "short" (one or two
> steps) beatpath to every candidate with a higher minimum
> pairwise score.
>

This is an interesting step, but I don't think it can be included without
some kind of philosophical motivation. And to keep it simple, I'd limit it
to one step. That is, if the ranking is A, B, C, D, E, E would be included
only if it requires only one indirect step to defeat each of A, B, C, and D.

(4).  Repeat step 2.  Then margins-sort the resulting scores and elect
> the highest-ordered candidate.*
>

If you're using the ERSf method above, scores would have to be
re-normalized before counting.


>
> This is trying to meet Clone Independence, Mono-raise, Chicken Dilemma,
> Non-Drastic Defense.
>

An interesting idea. But why wouldn't sorted margins after the first
re-scoring work?


> Chris Benham
>
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