[EM] Compromise between IRV and Condorcet methods

[Richard]

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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