<div dir="auto">Consider the ballot profile ...<div dir="auto"><br></div><div dir="auto">p A>B=C</div><div dir="auto">q B>A=C</div><div dir="auto">r C>B>A</div><div dir="auto"><br></div><div dir="auto">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.</div><div dir="auto"><br></div><div dir="auto">So geometrically the three factions form an isosceles triangle with the longest side opposite the second faction.</div><div dir="auto"><br></div><div dir="auto">So geometric proximity/affinity would require that candidate B be a clear second choice in the first and third factions.</div><div dir="auto"><br></div><div dir="auto">So the geometrically corrected (consistent) factions would have to be ...</div><div dir="auto"><br></div><div dir="auto">p A>B</div><div dir="auto">q B</div><div dir="auto">r C>B</div><div dir="auto"><br></div><div dir="auto">In that case B would have to be the CW if neither A nor B was a majority winner. </div><div dir="auto"><br></div><div dir="auto">In short, a Condorcet cycle would be inconsistent with the geometric affinities.</div><div dir="auto"><br></div><div dir="auto">Let's look at Kemeny-Young:</div><div dir="auto"><br></div><div dir="auto">The respective total distances from the three original factions to the respective six finish orders would be ....</div><div dir="auto"> </div><div dir="auto">p/2+1.5q+3r to ABC,</div><div dir="auto"><br></div><div dir="auto">p/2+2q+2r to ACB,</div><div dir="auto"><br></div><div dir="auto">1.5p+g/2+2r to BAC,</div><div dir="auto"><br></div><div dir="auto">2.5p+q/2+r to BCA,</div><div dir="auto"><br></div><div dir="auto">1.5p+2.5q+r to CAB</div><div dir="auto"><br></div><div dir="auto">1.5p+1.5q to CBA</div><div dir="auto"><br></div><div dir="auto">Suppose, for example p=q=30, and r=40.</div><div dir="auto"><br></div><div dir="auto">Then the respective total costs (distances) would be [check my arithmetic]...</div><div dir="auto">180 for ABC,</div><div dir="auto">155 for ACB,</div><div dir="auto">140 for BAC,</div><div dir="auto">130 for BCA,</div><div dir="auto">160 for CAB, and</div><div dir="auto">90 for CBA.</div><div dir="auto"><br></div><div dir="auto">So in this case CBA would be the K-Y finish order, contrary to the geometric affinities implied by the Kendall-tau geometry ...</div><div dir="auto"><br></div><div dir="auto"><div dir="auto" style="font-family:sans-serif">30 A>B</div><div dir="auto" style="font-family:sans-serif">30 B</div><div dir="auto" style="font-family:sans-serif">40 C>B,</div><div dir="auto" style="font-family:sans-serif"><br></div><div dir="auto" style="font-family:sans-serif">which would make B the Condorcet Winner, and BCA the fish order.</div><div dir="auto" style="font-family:sans-serif"><br></div><div dir="auto" style="font-family:sans-serif">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 function.</div><div dir="auto" style="font-family:sans-serif"><br></div><div dir="auto" style="font-family:sans-serif">That's one reason I reject K-Y ... the other reason is that the Kendall-tau metric is clone dependent.</div><div dir="auto" style="font-family:sans-serif"><br></div><div dir="auto" style="font-family:sans-serif">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).</div><div dir="auto" style="font-family:sans-serif"><br></div><div dir="auto" style="font-family:sans-serif">-Forest</div><div dir="auto" style="font-family:sans-serif"><br></div></div><br><br><div class="gmail_quote" dir="auto"><div dir="ltr" class="gmail_attr">El sáb., 12 de mar. de 2022 9:58 a. m., Forest Simmons <<a href="mailto:forest.simmons21@gmail.com">forest.simmons21@gmail.com</a>> escribió:<br></div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="auto"><div dir="auto">Kristofer,</div><div dir="auto"><br></div>45 ABC<div dir="auto">35 BCA</div><div dir="auto">25 CAB</div><div dir="auto"><br></div><div dir="auto">Each of the A and B factions has more than a third of the votes. Candidate A defeats B pairwise. </div><div dir="auto"><br></div><div dir="auto">Almost every respectable method except TACC (as well as most non-respectable methods) agree that candidate A should have the greatest winning probability.</div><div dir="auto"><br></div><div dir="auto">But some nagging doubt persists ... whence the Condorcet Cycle?</div><div dir="auto"><br></div><div dir="auto">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.</div><div dir="auto"><br></div><div dir="auto">And yet the voted ballots in our above three faction example give each candidate a turn at last place.</div><div dir="auto"><br></div><div dir="auto">Somebody's lowest preference is either mis-triangulated or mis-represented.</div><div dir="auto"><br></div><div dir="auto">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.</div><div dir="auto"><br></div><div dir="auto">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.</div><div dir="auto"><br></div><div dir="auto">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.</div><div dir="auto"><br></div><div dir="auto">So A's win very likely was achieved by burial of C, which in turn, is the likely true CW.</div><div dir="auto"><br></div><div dir="auto">So, is this enough evidence to convict A and elect C?</div><div dir="auto"><br></div><div dir="auto">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.</div><div dir="auto"><br></div><div dir="auto">So what test would give the compelling evidence? </div><div dir="auto"><br></div><div dir="auto">Only checking the B>C pairwise defeat has a chance of definitively ("dispositively") settling the case.</div><div dir="auto"><br></div><div dir="auto">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.)</div><div dir="auto"><br></div><div dir="auto">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.</div><div dir="auto"><br></div><div dir="auto">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)?</div><div dir="auto"><br></div><div dir="auto">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).</div><div dir="auto"><br></div><div dir="auto">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.</div><div dir="auto"><br></div><div dir="auto">Thoughts?</div><div dir="auto"><br></div><div dir="auto">-Forest</div><div dir="auto"><br></div><div dir="auto"><br></div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">El vie., 11 de mar. de 2022 2:20 a. m., Kristofer Munsterhjelm <<a href="mailto:km_elmet@t-online.de" target="_blank" rel="noreferrer">km_elmet@t-online.de</a>> escribió:<br></div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">On 3/11/22 6:04 AM, Kevin Venzke wrote:<br>
> Hi Kristofer,<br>
> <br>
>> On 3/11/22 12:33 AM, Kristofer Munsterhjelm wrote:<br>
>> Does it also apply to the generalization where you just take the two<br>
>> candidates with the most first preferences? I'm not sure.<br>
> <br>
> In the three-candidate case, electing the pairwise winner between the top two<br>
> candidates is basically IRV.<br>
> <br>
> Without the 1/3 limit it could happen that the FPW gets more votes and changes<br>
> who the second place candidate is. He might not beat the new one.<br>
> <br>
> This is interesting though. The "obvious" way to expand IFPP to many candidates<br>
> is to eliminate candidates with a below-average vote count. But it seems like<br>
> the 1/3 rule was the important thing, as it's what enforces that always either<br>
> one or two candidates are eligible to win, and these candidates can't be harmed<br>
> by getting more votes.<br>
<br>
Yes, that also explains where the "third" in dominant mutual third comes <br>
from. Like with Droop proportionatliy, it's the smallest quota so that <br>
only two candidates can exceed it. And that would also suggest that (at <br>
least by this approach), third is the best we can do; there's no, say, <br>
dominant mutual quarter for Condorcet.<br>
<br>
> Tricky, to reduce a scenario to this state without breaking mono-raise.<br>
<br>
We could of course just stitch something together, e.g. if there're <br>
exactly two candidates above 1/3 fpp, elect the candidate who pairwise <br>
beats the other, otherwise just elect the Plurality winner. This should <br>
be monotone because raising A doesn't harm A when A has >1/3 fpp, and <br>
raising B to >1/3 fpp gives him a second chance against A (if B beats A <br>
pairwise).<br>
<br>
It's not very elegant: the seams are very obvious. But perhaps elegance <br>
can come later... or perhaps it will be induced by turning DMT candidate <br>
BR into full DMTBR.<br>
<br>
-km<br>
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</blockquote></div>
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