[EM] “Monotonic” Binomial STV

[Kevin Venzke]

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.)