[EM] Duncan Proposal Draft

[Michael Ossipoff]

BTW, Forest has suggested a small modification of the CTE that I posted
here:

Instead of eliminating the lowest Borda-scorer, eliminate the pair-loser of
the 2 lowest Borda-scorers.

Even with only 3 candidates, that won’t elect BF unless CW is the highest
Borda-scorer.

If CW is the lowest Borda-scorer, which is most likely, Bus wins & the
burial is penalized.

With more candidates, & therefore likely more Buses, it becomes even less
likely for the winner to not be a Bus.

On Mon, Oct 23, 2023 at 06:01 C.Benham <cbenham at adam.com.au> wrote:

> Are you referring to how equal-ranking is counted in Borda?
>
> Yes, mainly truncated ballots.  Do you consider that truncation is the
> same thing as ranking equal bottom?
>
> I advocate
> that, in these methods’ Borda-count,  a ranking give each of its
> ranked-candidates a number of points equal to the number of candidates in
> the ranking who aren’t ranked over hir.
>
> Can you clarify the exact meaning of the phrase "in the ranking"?  Is this
> a hint that your answer to my last
> question is "no"?
>
> And why is that what you advocate?  In some online discussion I saw it was
> mentioned (I think by N.Tideman) that
> the Baldwin method doesn't meet Condorcet unless the Fractional (i.e.
> based on the symmetrically completed ballots)
> version of Borda count is used.
>
> So I've tended to assume that that is the "correct" way of doing Borda
> counts.
>
> Say there are three candidates A,B,C.   Say 46 ballots bullet-vote A.  In
> the fractional version these votes are counted the same as
> 23 A>B, 23 A>C,  giving 92 points to A and 23 points each to B and C.
>
> Is that the same as what you advocate, or would you have those truncating
> ballots give 2 points to A and zero points to both of B and C?
> Or something else?
>
> * So another way of putting it is:  "If there is no CW, elect the member of
> ** the Smith set with the second-worst score".
> *
> Would you mind telling why that’s another way of putting it?
>
>
> By definition the members of the Smith set all pairwise beat all the
> non-member candidates, and also have a pairwise defeat at the hands
> of one of the other members (assuming there are three members).
>
> So the Smith set member with the highest score must be pairwise beaten by
> one with a lower score. That much is clear.
>
> To be honest I may have confused myself again as to further details.  It
> seems to be possible for the lowest-scored Smith-set member to win.
>
> Say the top cycle is (in terms of score order)  Middle > High > Low >
> Middle.
>
> Then High is disqualified by being pairwise beaten by Middle and Middle is
> disqualified by being pairwise beaten by Low.
>
> So Low wins.
>
> Remember that sincere top-cycles are vanishingly rare.
>
>
> What mind-reading technology have you accessed to determine that?  I don't
> know how you can know that.
>
> Simple scenarios with lots of truncation and a top cycle look very
> plausible to me, so I don't know why I should
> accept your assurance that such a thing would be "vanishingly rare".
>
> Also it seems to me you are giving yourself a marketing problem:
>
> "Ok, we admit that Condorcet methods are very vulnerable to Burial
> strategy, so in a effort to combat that we have
> to employ this apparently nonsensical, anti-monotonic, anti-intuitive
> completion method.
>
> But don't worry, probably we'll never have to use it."
>
> Chris B.
>
>
>  *Michael Ossipoff* email9648742 at gmail.com
> <election-methods%40lists.electorama.com?Subject=Re%3A%20%5BEM%5D%20Duncan%20Proposal%20Draft&In-Reply-To=%3CCAOKDY5D-CUNWajrZxSi8VauLY8FAys8vnt6SvX%3D7KiQFeEpeJw%40mail.gmail.com%3E> *Sat
> Oct 21 12:38:58 PDT 2023*
>
>
> ------------------------------
>
> On Sat, Oct 21, 2023 at 11:05 C.Benham <cbenham at adam.com.au <http://lists.electorama.com/listinfo.cgi/election-methods-electorama.com>> wrote:
>
> "Am I right in assuming that the Borda counts are based on the symmetrically
> completed ballots?"
>
> Are you referring to how equal-ranking is counted in Borda? I advocate
> that, in these methods’ Borda-count,  a ranking give each of its
> ranked-candidates a number of points equal to the number of candidates in
> the ranking who aren’t ranked over hir.
>
>
>
> >* Duncan Definition:
> *>>* In the vast majority of the cases ... those in which the pairwise counts
> *>* of the ballots unambiguously identify the candidate that pairbeats each of
> *>* the others ... elect that candidate.
> *>>* Otherwise, elect the highest score candidate that pairbeats every
> *>* candidate with lower score.
> *>>>* So another way of putting it is:  "If there is no CW, elect the member of
> *>* the Smith set with the second-worst score".
> *>
> Would you mind telling why that’s another way of putting it?
>
>
> >>* To put it bluntly, that is bound to have monotonicity problems and doesn't
> *>* fly philosophically.
> *>
> How doesn’t it fly philosophically?
>
> >>>* Trying to deter or frustrate order-reversal Burial strategy is fine, but
> *>* the algorithm should "appear fair" and be able to be justified when
> *>* we assume that all the votes are sincere (or even just all equally likely
> *>* to be sincere).
> *>
> Remember that sincere top-cycles are vanishingly rare. That’s why
> Sequential-Pairwise’s Pareto failure isn’t important, & it’s why
> MinMax(wv)’s Condorcet Loser failure isn’t important.
>
> It matters much more what happens in a strategic cycle. Does the method
> reward or penalize burial?  …or neither?
>
> Duncan tends to often do neither, because it will likely disqualify both
> Bus & BF. (I defined those usages when I defined CTE.
>
> I prefer CTE to Duncan, because it more often penalizes, instead of merely
> not rewarding.
>
> But I remind you that these methods are intended to deter
> probabilisticallly, but aren’t claimed to penalize burial in every possible
> example.
>
> >>
>
> >>* That breaks (at least one version of) "Double Defeat".  B is pairwise
> *>* beaten by a candidate with a higher "score".
> *>>>* Chris B.
> *>>>* On 14/10/2023 4:43 am, Michael Ossipoff wrote:
> *>>* Yes, I like Duncan because burying the CW in an attempt to help your
> *>* favorite won’t help hir when it causes hir disqualification, as it probably
> *>* will.
> *>>* …& Duncan is remarkably briefly-defined, needing only a very slight
> *>* modification of Black’s method.
> *>
>
>
>
>
>
>
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