Ranked Robin: Difference between revisions

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If there is a tie (including [[Condorcet paradox|Condorcet cycles]]), use the '''1<sup>st</sup> Degree''' tie-breaking method to resolve it. If there is still a tie, use the '''2<sup>nd</sup> Degree''' tiebreaker, and so on.
 
'''1<sup>st</sup> Degree:''' Declare the tied candidates finalists. For each finalist, subtract the number of ballots on which they lost to each other finalist from the number of ballots on which they beat each other finalist. The finalist with the greatest total difference is elected. For example, let <math>A_w</math> be the number of ballots on which finalist <math display="inline">A</math> beats each other finalist (equivalent to the sum of the values in <math display="inline">A</math>'s row in a preference matrix consisting only of finalists) and let <math>A_l</math> be the number of ballots on which <math display="inline">A</math> loses to each other finalist (equivalent to the sum of the values in <math display="inline">A</math>'s column in a preference matrix consisting only of finalists); <math display="inline">A</math>'s total difference is <math>A_w-A_l</math>. This is mathematically equivalent to the [https://en.wikipedia.org/wiki/Borda_count#Tournament-style_counting_of_ties tournament-style of the Borda count] (among only the finalists), where candidates get, per ballot, 1 point for each candidate they beat and ½ point for each candidate they tie. TheThere are two alternate, more concise ways of describing the '''1<sup>st</sup> Degree''' tiebreaker.
 
* ''If there is a tie, then for each tied candidate, subtract the number of voters who prefer each other tied candidate from the number of voters with the opposite preference. Elect the tied candidate with the greatest total difference.''
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