[EM] “Monotonic” Binomial STV

[Kristofer Munsterhjelm]

On 27.02.2022 20:30, 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 probably is a special case of a more general method.

I think this special case can be further simplified by removing the
square root because the square root is monotone for nonnegative
integers: sqrt(x) > sqrt(y) iff x > y. And for the sake of simplicity,
the rule "elect the least keep value" can be replaced with "elect the
candidate with the greatest score" by flipping numerator and denominator.

So we have: keep value of A is sqrt( (quota/first prefs) * (last prefs /
quota))

which is proportional to sqrt(last prefs / first prefs)

which is monotonely related to (last prefs)/(first prefs)

and so in score terms, score(A) = fpA/lpA

where fpA is the number of first preferences for A, and lpA is the
number of last preferences for A, and the candidate with the highest
score is elected,

will elect the same candidates as (this special case of) Binomial STV does.

(You could try to add this method to the simulator we've been
discussing, I guess, and see where it ends up :-) Though it has the same
ambiguity as Antiplurality about what to do with truncation and equal-rank.)

> 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.

Yes; or a three-seat election with two winners and three candidates.

-km