[EM] Tie Breaking Version of Burial Resistant Method
voting at ukscientists.com
Thu Dec 30 10:43:38 PST 2021
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.
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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