[EM] Single-candidate DMTBR idea
Forest Simmons
forest.simmons21 at gmail.com
Sat Mar 12 09:58:40 PST 2022
Kristofer,
45 ABC
35 BCA
25 CAB
Each of the A and B factions has more than a third of the votes. Candidate
A defeats B pairwise.
Almost every respectable method except TACC (as well as most
non-respectable methods) agree that candidate A should have the greatest
winning probability.
But some nagging doubt persists ... whence the Condorcet Cycle?
A general scalene triangle has a longest side, and the endpoints of that
side are further from each other than they are from the vertex V opposite
that side, which means that the V faction favorite cannot be the rational,
sincere last choice of any of the three factions.
And yet the voted ballots in our above three faction example give each
candidate a turn at last place.
Somebody's lowest preference is either mis-triangulated or mis-represented.
Suppose the smallest faction, 25CAB, to be the inaccurate one ... with
true preferences 25CBA. Then B would be the true CW, and they would be
kicking themselves for inadvertently reversing their B>A preference for
their ballots.
On the other hand, suppose the suspicious burial arose from a CB to BC swap
in the largest faction. Then that faction would be congratulating itself
for its good fortune in converting their favorite A into a winner.
So, in the likely case that the ABCA ballot cycle was created artificially,
the only faction that actually improved its outcome by the burial is the
most likely culprit.
So A's win very likely was achieved by burial of C, which in turn, is the
likely true CW.
So, is this enough evidence to convict A and elect C?
No. This is like a Columbo episode where Columbo knows "who done it", on
the basis of compelling logic based on circumstantial evidence, but the
criminal is still taunting him for not having the kind of proof that will
hold up in court.
So what test would give the compelling evidence?
Only checking the B>C pairwise defeat has a chance of definitively
("dispositively") settling the case.
How can we check that supposed defeat without exhuming the victim's body
from the grave? (The Columbo equivalent of going back to the polls to check
the head-to-bead result between B and C.)
That's why we allow each voter an optional second ballot (paired with their
first) to be used in case (and only in case) a final binding, two-finalist
runoff is needed for definitive disposition of the election.
How would this work if the basic method were DMC (which nominally elects
the lowest implicit approval candidate that pairwise defeats every
candidate with greater IA)?
The two finalists are the nominal DMC winner W, and the candidate X that
would be the DMC winner if it pairwise defeated W, i.e. the candidate X
that defeats every candidate with greater IA, except possibly W itself (if
there is such an X).
It seems to me that this tweak of DMC would make it more resistant to
burial and chicken than any of the methods that elect A in our opening
example above ... better than any of the acceptable methods except TACC,
and even better than TACC because instead of merely punishing the burial of
C, it leads to vindication and election of C.
Thoughts?
-Forest
El vie., 11 de mar. de 2022 2:20 a. m., Kristofer Munsterhjelm <
km_elmet at t-online.de> escribió:
> On 3/11/22 6:04 AM, Kevin Venzke wrote:
> > Hi Kristofer,
> >
> >> On 3/11/22 12:33 AM, Kristofer Munsterhjelm wrote:
> >> Does it also apply to the generalization where you just take the two
> >> candidates with the most first preferences? I'm not sure.
> >
> > In the three-candidate case, electing the pairwise winner between the
> top two
> > candidates is basically IRV.
> >
> > Without the 1/3 limit it could happen that the FPW gets more votes and
> changes
> > who the second place candidate is. He might not beat the new one.
> >
> > This is interesting though. The "obvious" way to expand IFPP to many
> candidates
> > is to eliminate candidates with a below-average vote count. But it seems
> like
> > the 1/3 rule was the important thing, as it's what enforces that always
> either
> > one or two candidates are eligible to win, and these candidates can't be
> harmed
> > by getting more votes.
>
> Yes, that also explains where the "third" in dominant mutual third comes
> from. Like with Droop proportionatliy, it's the smallest quota so that
> only two candidates can exceed it. And that would also suggest that (at
> least by this approach), third is the best we can do; there's no, say,
> dominant mutual quarter for Condorcet.
>
> > Tricky, to reduce a scenario to this state without breaking mono-raise.
>
> We could of course just stitch something together, e.g. if there're
> exactly two candidates above 1/3 fpp, elect the candidate who pairwise
> beats the other, otherwise just elect the Plurality winner. This should
> be monotone because raising A doesn't harm A when A has >1/3 fpp, and
> raising B to >1/3 fpp gives him a second chance against A (if B beats A
> pairwise).
>
> It's not very elegant: the seams are very obvious. But perhaps elegance
> can come later... or perhaps it will be induced by turning DMT candidate
> BR into full DMTBR.
>
> -km
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