[EM] Tie Breaking Version of Burial Resistant Method

[Richard Lung]

James, good to hear from you again,
Reasons do exist for counting abstentions. 
Where the abstentions appear in a voters order of preferences influences the relative importance of votes meant to elect or exclude in a binomial count. A binomial count is where, say you have 10 candidates seeking 5 seats. Preferences 1 to 5 will more or less help elect 5 candidates. Preferences 6 to 10 will less or more help to exclude 5 candidates.
Where the abstentions occur in voters orders of preference will decide the relative importance voters give to electing or excluding candidates.

Another reason for counting abstentions, with a binomial count, is that if they add up to a quota, one seat will remain unfilled. This is an incentive for organisations to nominate better candidates, rather than stooges, so no seats are left unfilled. 

I don't see lower preferences as merely contingent but as also positively wanting several representatives.


Regards,
Richard Lung.



On 1 Jan 2022, at 1:45 pm, James Gilmour <jgilmour at globalnet.co.uk> wrote:

Richard
I think there is a difference of interpretation here about the "meaning" of incomplete preferences  -  and about not voting.

"Election method (system) is about representing all voters", indeed.  Note your word "voters".  So those electors who choose to opt out by not voting (abstain) are themselves making a conscious decision not to take part is the process of choosing the representatives.  I can see nothing that implies abstentions must be counted.

Similarly, in an STV-PR election I always understood that any second and subsequent preferences indicated contingency choices that would come into play only if your first choice candidate was excluded (because too few voters supported him) or was already elected (and so did not need your support, and could not in due proportion also represent you).  When a voter does not mark a preference against every candidates on her ballot paper, she is telling the Returning Officer to give effect to the preferences she has marked, and then said: "If it comes to a choice among any of the remaining candidates (whom I have not marked), I am not able or don't wish to express any preference among them and I am happy to leave any choice that has to be made among those candidates to those voters who do have preferences among those candidates.

James Gilmour


> -----Original Message-----
> From: Election-Methods [mailto:election-methods-bounces at lists.electorama.com] On Behalf Of Richard Lung
> Sent: 30 December 2021 18:44
> To: Forest Simmons <forest.simmons21 at gmail.com>; EM <Election-methods at lists.electorama.com>
> Subject: Re: [EM] Tie Breaking Version of Burial Resistant Method
> 
> Election method is about representing all voters, not about beating other candidates. A frequent objection to the latter
> approach is that in ruling out candidates, voter information about them is being lost.
> 
> The key to progress in election method is to count all the preferential information supplied by the voters. That means not
> only counting votes that help to elect, but counting votes that help to exclude. Hence, my binomial count. Which also
> implies the abstentions must be counted. --  To properly balance the books, all preferences must be counted.
> 
> Regards,
> Richard Lung.
> 
> 
>> On 30/12/2021 00:00, Forest Simmons wrote:
>> Let's say a candidate is at the "bottom" of a set of candidates if it
>> is not ranked above even one member of the set.
>> 
>> The basic method is to elect the candidate who on the most ballots
>> pairwise beats every bottom candidate.
>> 
>> Here's the procedure Proc(Beta) with built in tie breaker:
>> 
>> Beta is the set of ballots serving as input for this election
>> procedure. Let K be the set of candidates ranked by the ballots of Beta.
>> 
>> For each candidate X in K let n(X) be the number of ballots B in Beta
>> on which X pairwise beats every bottom candidate of B.
>> 
>> Let T be the set argmax(n(X)), i.e.the tied winning set. If T has only
>> one member, elect that member. Else if T=K, i.e. all candidates are in
>> an exact tie, then elect by random ballot from K. Otherwise, elect
>> Proc(Beta(T)), i.e. apply this procedure recursively to the set of
>> ballots restricted to T.
>> 
>> Without the tiebreaking extension, this method is about as susceptible
>> to ties as Implicit Approval.
>> 
>> Simple random ballot would be adequate for this single winner method,
>> but to get an "order of finish" method based on the function
>> X---->n(X) the recursive version would be very appropriate .... by
>> transforming n(X) into a polynomial in powers of epsilon with an
>> additional term for each recursive pass through the procedure.


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