[EM] "Independence of cycles" and a possible new method.
forest.simmons21 at gmail.com
Sun Dec 12 23:35:07 PST 2021
For what it's worth, three way tie of the opposite cyclic order would
cancel the above tied cycle and reverse it for x>2.
x: C>B> A
A perfectly balanced cycle is not neutral in its influence ... it exerts
The torsion from the original "tied" cycle reinforces the smaller faction
order, but goes against the order of the larger faction ... weakening its
If we were electing cyclic social orders instead of individual candidates
this would make more sense ... add two of these neutral cycles together ...
if they were absolutely neutral they wouldn't be able to affect each other.
Let's have an election to decide democratically if rock/paper/scissors is
the true cyclic order or its reverse. Some people may believe that rocks
are stronger than paper and that paper can cover scissors and that scissors
can chip away at rocks, for example.
El dom., 12 de dic. de 2021 3:38 a. m., Kristofer Munsterhjelm <
km_elmet at t-online.de> escribió:
> On 12/12/21 11:04 AM, Toby Pereira wrote:
> > If you look at the example here:
> > https://www.rangevoting.org/TobyCondParadox.html
> > <https://www.rangevoting.org/TobyCondParadox.html> having a tie cycle
> > a Condorcet election can change the result. The ballots are essentially:
> > 1 voter: A>B>C
> > 2 voters: B>A>C
> > Plus
> > 2 voters: A>B>C
> > 2 voters: B>C>A
> > 2 voters: C>A>B
> > A is the Condorcet winner in this election, beating both B and C by 5
> > votes to 4. However, in the top section of ballots, B is the Condorcet
> > winner, and the bottom section is a three-way tie as it is a perfect
> > cycle. Adding in tie cycles to change the result could be seen as a
> > failure of "independence of cycles" (or perhaps there is a better name).
> > This is also a specific case of failure of consistency. Or maybe a weak
> > failure because B doesn't win both the sub-elections. B would have an
> > outright win from the top ballots and tie from the bottom ballots, but
> > loses to A when all are combined.
> > This example, by the way, is a simplified version of something that
> > Donald Saari used to promote the Borda Count.
> > The problem is that in an election with more than three candidates, you
> > couldn't simply remove the cycles and calculate the result. Ballots and
> > candidates would potentially be involved in many intertwined cycles, so
> > there would be no straightforward way of doing it.
> > But what you can do is compare every possible triplet of candidates
> > (like Condorcet methods compare pairs). For each triplet, all tie cycles
> > are removed and you look at the head-to-heads.
> You could do this, but as I understand the example, the method would no
> longer be a Condorcet method. You could also define the irrelevance of
> cycles criterion, perhaps something like:
> Removing a constant number of voters who together form an exact tied
> Condorcet cycle should not modify the output.
> Though I'm not sure what the implications would be - or if it's possible
> to pass by any method that fails IIA.
> As for using triplets instead of pairs, I think Stensholt suggested that
> doing so might be a way to generalize his BPW method, which is only
> defined for three candiates. Similarly, it might be a way of
> generalizing my fpA-fpC, though I'm again unsure how to do so and
> preserve the desired properties of DMTBR and monotonicity.
> Election-Methods mailing list - see https://electorama.com/em for list
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