[EM] Compromise between IRV and Condorcet methods

[Kevin Venzke]

 Hi Chris,

Here's a scenario with a difference:

61: A>B>C
53: B
40: D>C
39: C

Benham (and IRV) eliminate C first, and eventually elect A.

Richard's RCIPE method eliminates D first, and then C goes on to win. This is
also the Smith//IRV treatment, in this case.

I agree that it's difficult to mix IRV with another method and retain IRV's
satisfied criteria.

Kevin
votingmethods.net


Chris Benham via Election-Methods <election-methods at lists.electorama.com> a écrit :
> 
> Richard,
> 
> As I just commented in my reply to Kevin,  I think Hare makes a bad a
> mixer and so is difficult to fruitfully "refine".
> 
> Can you give an example where your suggested method performs better than
> plan Hare (aka IRV)?
> 
> I think your suggested method would be quite a bit more difficult to
> hand-count than say Benham. With Benham when considering the candidate
> that Hare would next eliminate we only have to establish that it has a
> single pairwise defeat (before eliminating it), not that it loses all
> its pairwise contests.
> 
> Can you give an example where your method gives a better (or just
> different) result than Benham?
> 
> Chris
> 
> 
> On 22/08/2025 6:47 am, Richard via Election-Methods wrote:
> > On 8/21/25 11:37, Chris Benham via Election-Methods wrote:
> > > *Elect whichever of the Hare winner and the most approved candidate
> > > pairwise beats the other.*
> >
> > Here I'll put in a plug for refining IRV by eliminating pairwise
> > losing candidates when they occur.  It's a simple compromise between
> > IRV and Condorcet methods that isn't "clunky" and yields lots of "bang
> > for the buck."
> >
> > A pairwise losing candidate is a candidate who loses every one-on-one
> > contest against every other remaining candidate.
> >
> > Only when a counting round lacks a pairwise losing candidate does the
> > combined method fall back on eliminating the candidate with the fewest
> > transferred votes.
> >
> > Richard Fobes