Ranked Robin: Difference between revisions

'''1^st Degree:''' Declare the tied candidates finalists. For each finalist, subtract the number of ballotsvotes on which they lost topreferring each other finalist from the number of ballots onvotes whichpreferring theythem beatover each other finalist. The finalist with the greatest total difference is elected. For example, let <math>A_w</math> be the number of ballotsvotes on whichpreferring finalist <math display="inline">A</math> beatsover 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 ballotsvotes onpreferring whicheach other finalist over <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. There are two alternate, more concise ways of describing the '''1^st Degree''' tiebreaker.

* ''If there is a tie, then for each tied candidate, subtract the number of votersvotes who preferpreferring each other tied candidate from the number of votersvotes with the opposite preference. Elect the tied candidate with the greatest total difference.''

* ''Among the candidates who tie for winning the most head-to-head matchups, elect the tied candidate with the best average rank.''

'''2^nd Degree:''' For each tied finalist, subtract the number of ballotsvotes on which they lost topreferring each other candidate (including candidates who are not finalists) from the number of ballots onvotes whichpreferring theythem beatover each other candidate (including candidates who are not finalists). The tied finalist with the greatest total difference is elected.

'''4^th Degree:''' If there is still a tie after the '''3^rd Degree''' tiebreaker, it is unlikely that the '''4^th Degree''' tiebreaker will break that tie, as it will only work if the tied candidates have matchup losses against other candidates. Find the shortest [[beatpath]] from each tied candidate to each other tied candidate. For each tied candidate, for each shortest beatpath to another tied candidate, for each pairwise victory in the beatpath, subtract the number of ballotsvotes on whichpreferring the losing candidate was preferred over the winning candidate from the number of ballotsvotes on whichpreferring the winning candidate was preferred over the losing candidate. Sum these differences within each selected beatpath to get the total strength of each selected beatpath. Sum each tied candidate's total beatpath strengths over other tied candidates. Elect the tied candidate with the greatest sum of beatpath strengths. If there are multiple shortest beatpaths from one tied candidate to another, select the one with the lowest total strength.

'''4^th Degree:''' The shortest beatpath from Ava to Bianca is Ava→Deegan→Bianca and the shortest beatpath from Bianca to Ava is Bianca→Cedric→Ava. The difference between the number of votersvotes who preferpreferring Ava over Deegan and the number of voters whovotes preferpreferring Deegan over Ava is <math>39-38=1</math>. From Deegan to Bianca, the difference is <math>37-31=6</math>. The sum of the differences in the beatpath from Ava to Bianca (the total beatpath strength) is <math>1+6=7</math>. From Bianca to Cedric, the difference is <math>35-28=7</math>. From Cedric to Ava, the difference is <math>33-26=7</math>. The total beatpath strength from Bianca to Ava is <math>7+7=14</math>. Bianca has the greatest (sum of) total beatpath strength(s) among tied candidates, so Bianca is elected.

If there is a [[Condorcet winner criterion|Condorcet Winner]], then simply show each of the winner's pairwise matchups against other candidates. This can either be shown as percentage of total votes for each candidate in a given pairwise matchup, or as the percentage point difference in favor of the winner if there's a desire to show less information. If there is no Condorcet Winner but a single candidate wins without any tie-breaker, show how many matchups each candidate won in addition to each of the winner's pairwise matchups. These two scenarios will cover the vast majority of real-world elections.

</blockquote>In the rare case of a '''2^nd Degree''' tie, if there are many candidates, it is recommended to focus on who the tied finalists are and their Total Advantages over all other candidates (which likely will not sum to 0).