[EM] "we only get one shot" (Re: RCV Challenge)

[Forest Simmons]

Kristofer and All,

What I get from this is to use a Coombs/STV hybrid. Perhaps mostly Coombs
for Elimination and STV quotas for election ... something like that.

El mar., 4 de ene. de 2022 2:55 p. m., Kristofer Munsterhjelm <
km_elmet at t-online.de> escribió:

> On 04.01.2022 21:02, Richard Lung wrote:
> >
> > KM,
> >
> > I suspect those comments stem from (an unsurprising) unfamiliarity with
> > binomial stv. It has an exclusion count but it does not exclude or
> > eliminate candidates, during the binomial count (count of elections and
> > count of exclusions, before an over-all deciding count) and so does not
> > fall foul of irrationalities from "premature exclusion" etc.
>
> Well, when you're saying that:
>
> > By the way, the point is that an election method should make use of
> > an exclusion count, as well as an election count.
> You're saying something about what's desirable for any election method,
> not just binomial STV. My point is that the very concept "exclusion
> count" might not even make sense in methods that are sufficiently
> different from binomial STV.
>
> I might be misunderstanding the concept of just what an exclusion count
> is, hence my question. But if the concept is to be applicable in full
> generality, so that we can say that a method with an exclusion count is
> better than one that isn't, then there must be some way of unambiguously
> determining whether some arbitrary given election method has an
> exclusion count or not. And I don't quite understand how that is to be
> decided.
>
> In other words, suppose that someone gives me an election method (like
> FPTP or Borda or Kemeny or Approval). How do I determine if it has an
> exclusion count?
>
> -km
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