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@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
On 8/22/25 01:31, Chris Benham wrote:
I think your suggested method would be quite a bit more difficult to
hand-count ....
The pairwise counts do not need to be counted at each elimination round.
Instead, determining which candidate wins each one-on-one contest is
done only once at the beginning. Then, during each round, just refer to
the list of which candidate wins each pairwise contest.
(Specifically this can be done by counting how many pairwise losses
occur for each candidate. The count that is one less than the number of
remaining candidates reveals which candidate, if any, is the pairwise
losing candidate.)
If the counting were done on stage in front of an audience, a table with
6 people counting paper ballots would be sufficient to do the pairwise
counting among the four candidates who have the greatest chance of
winning. One other person can put the paper ballots into stacks that
track transferred votes. Those stacks do not need to be looked at to
identify pairwise losing candidates.
Can you give an example where your method gives a better (or just
different) result than Benham?
Kevin supplied an example of this difference.
Regarding the desire for "better" results, instead of just getting
better results for a specific few scenarios, it's more meaningful to
measure how often the method fails each criterion.
Interestingly Kristofer's measurements about vulnerability to strategic
voting reveal that this RCIPE (Ranked Choice Including Pairwise
Elimination) method has a very low vulnerability that's similar to the
small vulnerability of Benham's method.
Very significantly, partly as a result of propaganda from IRV promoters,
lots of voters do not trust that the Condorcet winner always deserves to
win. In contrast, I've never heard of anyone thinking that eliminating
a pairwise losing candidate would be unfair.
Richard Fobes
On 8/22/25 01:31, Chris Benham wrote:
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
On 8/21/25 11:37, Chris Benham via Election-Methods wrote:
Kevin,
Thanks for that demonstration.
A much more simple method (using the same type of ballots) definitely
does meet Mono-add-Top:
Elect whichever of the Hare winner and the most approved candidate
pairwise beats the other.
James Green-Armytage mentioned a while ago that he thought that would
be a good method. At the time I had different priorities and
dismissed it as something clunky that fails Condorcet and Mono-raise,
but now I agree. As a practical proposition it is probably doubtful
that the extra complication versus plain Hare gives enough bang for
buck, and I suppose as well as failing Condorcet it fails Double
Defeat. But nonetheless it must be quite a bit more Condorcet
efficient than Hare, while hanging on to Mono-add-Top compliance.
Chris
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