[EM] RCIPE version 2
Kristofer Munsterhjelm
km_elmet at t-online.de
Thu Jul 22 06:04:57 PDT 2021
On 7/21/21 5:25 PM, VoteFair wrote:
> On 7/21/2021 4:41 AM, Kristofer Munsterhjelm wrote:
> > I would of course suggest moving the tiebreaker up here to replace the
> > current elimination condition :-)
> >
> > "Eliminate the candidate who has the smallest one-on-one pairwise count
> > against any other remaining candidate."
>
> I tried placing a higher priority on this kind of elimination because it
> avoids eliminating the Condorcet loser. However it causes odd and
> undesirable behavior regarding Clone Independence and IIA failures.
>
> Using IRV as the tie breaker protects against clone independence
> failures -- which means it protects against money-based vote splitting
> tactics.
I meant as a replacement for the pairwise support count step, which is
not cloneproof either. Since it's a sum of the number of votes, you
would be very likely to get a unique result, so the IRV and minmax steps
would very rarely come into play.
> > You need the pairwise matrix to determine the Condorcet loser anyway, so
> > there's not much of a gain in complexity.
>
> A major goal of this redesign is to make the method easier to
> demonstrate using an animated video.
>
> Introducing the idea of a pairwise matrix would not work well in an
> animated video. It's much too difficult for most voters to understand.
>
> Instead, this method can be animated by having upward-pointing arrows
> emerge from the marks on a ranked-choice ballot.
>
> For each candidate, the number of those arrows equals the number of
> marks to the right of the mark for that candidate. This assumes that
> unmarked candidates are marked by default at the far-right, lowest
> ranking level.
>
> I'm intending to use candidate names that are readily associated with a
> single letter, such as:
>
> Jay (J), Kay (K), Cici (C), Gigi (G)
>
> Each arrow will have the pairwise winner's initial at the top of the
> arrow and the pairwise loser's initial at the bottom of the arrow.
>
> That will allow:
>
> * Removing arrows that are associated with the candidate who was just
> eliminated
>
> * Using the same arrows both for checking for Condorcet losers -- when
> the arrows are grouped into subgroups (one group per candidate, one
> subgroup per arrow letter pair) -- and checking for the smallest
> pairwise support count -- when the arrows are not subgrouped. >
> I'm learning that what's easy to describe in words is not necessarily
> easy to explain using graphics.
I see, so you are in effect restricted to the sign-values of the
Condorcet matrix entries.
How about this?
- Eliminate the candidate with the least number of winning subgroups.
- If there is a tie, break that tie by IRV.
That's simpler and always eliminates the Condorcet loser (because such a
loser obviously has no winning subgroups at all). Because the winning
subgroups depend only on the number of candidates and not on the number
of voters, you should get ties more often, which means that you have a
greater chance of using the clone-independent component (IRV). And you
don't have to introduce un-subgrouped arrows at all.
Alternatively you could do:
- Eliminate the Condorcet loser
- If there is none, eliminate the candidate with the least number of
winning subgroups,
- If there is a tie, break it by ungrouped arrows,
- If there's still a tie, break it by IRV,
- If there's still a tie, break it by minmax.
But again, the ungrouped mechanic is not cloneproof.
-km
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