[EM] “Monotonic” Binomial STV

[Forest Simmons]

"...It follows that if the abstentions add up to a
 quota, a seat is not taken...."

Kind of like NOTA ... none of the above.

I'm trying to think how I would design a method in the spirit of Binomial
STV .... elections vs exclusions ... preferences vs reverse preferences.

Perhaps some variant of Bucklin that gradually collapses ballot rankings
inward (ER Whole?) when not enough top or bottom votes exist to meet quotas
for further inclusion or exclusion ... taking special care to insure both
monotonicity and clone independence in the process, if possible.

I think collapsing has more potential for monotonicity than does
elimination, and I'm glad that Binomial stv keeps all of the players in the
game until the final count, like Bucklin does.

-Forest



El dom., 27 de feb. de 2022 4:54 p. m., Richard Lung <
voting at ukscientists.com> escribió:

>
> On 28/02/2022 00:45, Richard Lung wrote:
> >
> >  Thanks for your thoughts, Kevin,
> >
> > In this simple instance, the election and exclusion quotas cancel. But
> > I would be lost without it, in multi-member PR cases of involved
> > transferable voting. There are a few examples in my e-books, (The
> > Super-Vote supercharged..., Elect and Exclude..., FAB STV...) free
> > from Smashwords, in epub format, and pdf versions free from archive.
> > org where putting "Richard Lung" in quotes in the text box should come
> > up with about 19 titles.
> >
> > The square root may not be strictly necessary, which may be why I keep
> > forgetting it. But it keeps the average keep values on a par with the
> > election and exclusion keep values. The square root is for the correct
> > form of the geometric mean, -- an important average.
> >
> > Yes, you are right, there is some other rule not stated -- All the
> > abstentions are counted. in more complex elections, they have to be,
> > so as not to distort the relative importnce of the election and
> > exclusion counts. It follows that if the abstentions add up to a
> > quota, a seat is not taken. This provides an incentive to nominate
> > good candidates, who work for the voters rather than their nominees.
> >
> > So, a candidate is not necessarily electable. More-over a large enough
> > quota like Hare, with a small number of seats would also be
> > prohibitive of election, given the voters free choice.
> >
> > Regards,
> >
> > Richard Lung.
> >
> >
> >
> > On 27/02/2022 19:30, Kevin Venzke wrote:
> >> Hi Kristofer/Richard,
> >>
> >> I wonder not just about the square root, but also if the quota has some
> additional
> >> role in the method, perhaps when there are 4+ candidates.
> >>
> >> Because this expression:
> >> ( quota / keep ) * ( exclude / quota )
> >> Appears to simplify to:
> >> ( exclude / keep )
> >>
> >> This creates the appearance that the quota has no effect on the outcome.
> >>
> >> Richard stated that final values below unity are electable. It looks
> like there
> >> will always be an electable candidate, unless it's a complete tie, or
> perhaps if
> >> there is some other rule not yet stated here.
> >>
> >> It seems to me that the 3-candidate 1-winner case of this method is
> monotone.
> >> It would help to see a four-candidate election resolved, too.
> >>
> >> Kevin
> >>
> >> Le dimanche 27 février 2022, 07:41:20 UTC−6, Kristofer Munsterhjelm<
> km_elmet at t-online.de>  a écrit :
> >>> On 27.02.2022 14:04, Richard Lung wrote:
> >>>> Thank you, Kristofer,
> >>>>
> >>>>
> >>>> for first example.
> >>>>
> >>>> The quota is 100/(1+1) = 50.
> >>>>
> >>>> Election keep value is quota/(candidates preference votes)
> >>>>
> >>>> Exclusion keep value equals quota/(candidates reverse preference
> vote):
> >>>>
> >>>> Geometric mean keep value ( election keep value multiplied by inverse
> >>>> exclusion keep value):
> >>> Geometric mean:
> >>>
> >>> A: square root of (50/51 x 2/50) ~ 0.198
> >>> B: square root of (50/49 x 1/50) ~ 0.143
> >>> C: ~= infinity (or very high)
> >>>
> >>> So B wins, having the lowest keep value. Is this correct?
> >>>
> >>> (You seem to have omitted the square root in your calculations, but it
> >>> shouldn't make a difference. Without the square root, A and B's values
> >>> are 0.0392 and 0.0204 respectively.)
> >> Hi Kristofer/Richard,
> >>
> >> I wonder not just about the square root, but also if the quota has some
> additional
> >> role in the method, perhaps when there are 4+ candidates.
> >>
> >> Because this expression:
> >> ( quota / keep ) * ( exclude / quota )
> >> Appears to simplify to:
> >> ( exclude / keep )
> >>
> >> This creates the appearance that the quota has no effect on the outcome.
> >>
> >> Richard says final values below unity are electable. It seems like
> there will
> >> always be an electable candidate, unless it's a complete tie, or
> perhaps if there
> >> is some other rule not yet stated here.
> >>
> >> Kevin
> >>
> >>
> >> Le dimanche 27 février 2022, 07:41:20 UTC−6, Kristofer Munsterhjelm<
> km_elmet at t-online.de>  a écrit :
> >>> On 27.02.2022 14:04, Richard Lung wrote:
> >>>> Thank you, Kristofer,
> >>>>
> >>>>
> >>>> for first example.
> >>>>
> >>>> The quota is 100/(1+1) = 50.
> >>>>
> >>>> Election keep value is quota/(candidates preference votes)
> >>>>
> >>>> Exclusion keep value equals quota/(candidates reverse preference
> vote):
> >>>>
> >>>> Geometric mean keep value ( election keep value multiplied by inverse
> >>>> exclusion keep value):
> >>> Geometric mean:
> >>>
> >>> A: square root of (50/51 x 2/50) ~ 0.198
> >>> B: square root of (50/49 x 1/50) ~ 0.143
> >>> C: ~= infinity (or very high)
> >>>
> >>> So B wins, having the lowest keep value. Is this correct?
> >>>
> >>> (You seem to have omitted the square root in your calculations, but it
> >>> shouldn't make a difference. Without the square root, A and B's values
> >>> are 0.0392 and 0.0204 respectively.)
> ----
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