[EM] Inclusion/Exclusion Multiwinner Approval Idea

Richard Lung voting at ukscientists.com
Mon Jan 10 10:49:03 PST 2022


Thank you, Forest,
Your scores appear to be what statisticians call weighting in a series progression. The Borda count is an example of weighting in arithmetic progression -- as first proposed by Borda.
Statisticians prefer weighting in arithmetic proportion to give the intervals of a range of data their true, rather than estimated, importance, if supplied in the data. The Gregory method or Senatorial rules of transferable voting are an example of this standard statistical technique.
Keeping track of vote transfers, by candidates keep values, is a very convenient version of this technique, I borrowed and extended from Meek method.

Regards,
Richard Lung.




On 6 Jan 2022, at 9:37 pm, Forest Simmons <forest.simmons21 at gmail.com> wrote:

Here's an idea that has nothing to do with Binomial STV except that it popped into my head while puzzling over clues patiently provided by Richard Lung:

Voted ballots mark some candidates as definitely approved, others as definitely disapproved, with the unmarked considered to be indifferent "abstentions."

Each possible slate of candidates is given both an inclusion score and an exclusion score by each ballot. The inclusion score is the sum 1+1/2+...1/j, where j is the number of slate members marked definitely approved by the ballot. Similarly, the exclusion score is the harmonic series truncated at the k_th term, where k is the number of definitely disapproved candidates excluded from the slate.

For each slate, both ballot scores (inclusive and exclusive) are summed over all ballots.

Finally, the geometric mean of the two sums becomes the final score for the slate under consideration.

The slate with the highest score is elected.

Those familiar with proportional approval voting (PAV), will recognize this method as a combination of ordinary PAV applied to both the slate and its complement.

There are only nCm possible slates of m candidates chosen from among n candidates, so not exponentially hard.

For manual counts you would need to stick to nominated slates.



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