[EM] Duncan Proposal Draft

[C.Benham]

> 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
>   <mailto: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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