[EM] Revised: Instant Pairwise Elimination (IPE)

[VoteFair]

On 1/13/2020 7:56 AM, C.Benham wrote:
 > I missed the earlier discussion on this.
 > ...
 > And what exactly is a "pairwise-losing candidate"?

My apologies. For brevity I omitted the first paragraph of the 
description, which is here:

"Instant Pairwise Elimination (IPE) eliminates one candidate at a time. 
During each elimination round the candidate who loses every pairwise 
contest against every other not-yet-eliminated candidate is eliminated. 
The last remaining candidate wins."

Hopefully now the second paragraph will make sense:

"If an elimination round has no pairwise-losing candidate, then the 
method eliminates the candidate with the largest pairwise opposition 
count, which is determined by counting on each ballot the number of 
not-yet-eliminated candidates who are ranked above that candidate, and 
adding those numbers across all the ballots. If there is a tie for the 
largest pairwise opposition count, the method eliminates the candidate 
with the smallest pairwise support count, which similarly counts support 
rather than opposition. If there is also a tie for the smallest pairwise 
support count, then those candidates are tied and all those tied 
candidates are eliminated in the same elimination round."

 > Just curious.

Curiosity is what led me to election-method reform. Thanks for asking.

Richard Fobes

On 1/13/2020 7:56 AM, C.Benham wrote:
> I missed the earlier discussion on this.
>
> So what if an elimination round /does /have a "pairwise-losing
> candidate", what then?
>
> And what exactly is a "pairwise-losing candidate"?
>
> Just curious.
>
> Chris Benham
>
> On 13/01/2020 12:14 pm, VoteFair wrote:
>> Based on a suggestion from a user on Reddit, I have revised the
>> definition of the Instant Pairwise Elimination method that previously
>> I published at Democracy Chronicles and then discussed here.
>>
>> The method still successively eliminates pairwise (Condorcet) losers.
>>
>> Now, instead of resolving Condorcet (rock-paper-scissors) cycles using
>> an "upside-down" version of instant-runoff voting (IRV), it uses
>> pairwise counts as described here:
>>
>> "If an elimination round has no pairwise-losing candidate, then the
>> method eliminates the candidate with the largest pairwise opposition
>> count, which is determined by counting on each ballot the number of
>> not-yet-eliminated candidates who are ranked above that candidate, and
>> adding those numbers across all the ballots. If there is a tie for the
>> largest pairwise opposition count, the method eliminates the candidate
>> with the smallest pairwise support count, which similarly counts
>> support rather than opposition. If there is also a tie for the
>> smallest pairwise support count, then those candidates are tied and
>> all those tied candidates are eliminated in the same elimination round."
>>
>> Below are my guesses for which fairness criteria it fails and passes.
>> Please tell me which guesses are not correct.
>>
>>   Condorcet: fail
>>   Condorcet loser: pass
>>   Ranks equal: pass
>>   Ranks greater than 2: pass
>>   Polytime: pass
>>   Resolvable: pass
>>   Majority: fail
>>   Majority loser: fail
>>   Mutual majority: fail
>>   Smith/ISDA: fail
>>   LIIA: fail
>>   IIA: fail
>>   Cloneproof: fail
>>   Monotone: fail
>>   Consistency: fail
>>   Reversal symmetry: fail
>>   Later no harm: fail
>>   Later no help: fail
>>   Burying: fail
>>   Participation: fail ?
>>   No favorite betrayal: fail ?
>>   Summable: O(N!) ?
>>
>> As I've said many times, it's the frequency with which the failures
>> occur that is much, much more important than simply counting how many
>> criteria it fails. I suspect that its frequencies of failure will be
>> quite low compared to most other single-winner methods, and may
>> approach the low frequencies that I believe characterize the
>> Condorcet-Kemeny method.
>>
>> I've created a page for this method on Electowiki. You are welcome to
>> edit that page with any corrections.
>>
>> BTW, I realize that it's possible that the alternate elimination
>> method always identifies the pairwise/Condorcet loser (if there is
>> one). If so, this would mean that the description could be
>> "simplified" to a single step (actually two steps in case there is a
>> tie). However, for the benefit of most voters who are not comfortable
>> with mathematics it's important to explicitly state that the first
>> priority is to eliminate the pairwise loser.
>>
>> Of course software that implements the method would do the
>> calculations using a much faster method than the counting method
>> described above. The description above is written to be understandable
>> to people who are not already familiar with pairwise counting.
>>
>> In advance, thank you for any feedback.
>>
>> Richard Fobes
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>
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