[EM] Geometric Inconsistency of Kemeny-Young (was Single-candidate DMTBR idea)

Forest Simmons forest.simmons21 at gmail.com
Sat Mar 12 15:03:27 PST 2022

Consider the ballot profile ...

p A>B=C
q B>A=C
r  C>B>A

Each ballot in the first faction has a Kendall-tau distance of 1.5 from
each ballot in the second faction and 2.5 from each ballot in the third
faction, while each ballot in the second faction has a distance of 1.5 from
each ballot in the third faction.

So geometrically the three factions form an isosceles triangle with the
longest side opposite the second faction.

So geometric proximity/affinity would require that candidate B be a clear
second choice in the first and third factions.

So the  geometrically corrected (consistent) factions would have to be ...

p A>B
q B
r C>B

In that case B would have to be the CW if neither A nor B was a majority

In short, a Condorcet cycle would be inconsistent with the geometric

Let's look at Kemeny-Young:

The respective total distances from the three original factions to the
respective six finish orders would be ....

p/2+1.5q+3r to ABC,

p/2+2q+2r to ACB,

1.5p+g/2+2r to BAC,

2.5p+q/2+r to BCA,

1.5p+2.5q+r to CAB

1.5p+1.5q  to CBA

Suppose, for example p=q=30, and r=40.

Then the respective total costs (distances) would be [check my
180 for ABC,
155 for ACB,
140 for BAC,
130 for BCA,
160 for CAB, and
90 for CBA.

So in this case CBA would be the K-Y finish order, contrary to the
geometric affinities implied by the Kendall-tau geometry ...

30 A>B
30 B
40 C>B,

which would make B the Condorcet Winner, and BCA the fish order.

So ordinary K-Y, which is based on the Kendall-tau metric, yields a result
inconsistent with preferences based on that same Kendal-tau distance

That's one reason I reject K-Y ... the other reason is that the Kendall-tau
metric is clone dependent.

Those are the reasons I have suggested a de-cloned version of Kendall-tau
together with some geometrically consistent applications of that new metric
(other than de-cloned K-Y).


El sáb., 12 de mar. de 2022 9:58 a. m., Forest Simmons <
forest.simmons21 at gmail.com> escribió:

> 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
>> ----
>> Election-Methods mailing list - see https://electorama.com/em for list
>> info
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