James Green-Armytage

Criteria passed and failed by Cardinal-Weighted Pairwise

 

            For the sake of convenience in this section alone, let me assume that the method being evaluated is cardinal pairwise, completed by beatpath[1] (in step 3 of the tally), without any of the additional provisions.

            Let’s also assume that our tie-breaking method is as follows: Create a tie-breaking ranking of the options (TBRO)[2] by choosing a random ballot and engraving all of the pairwise preferences expressed in that ballot onto the TBRO. Pairs that are not yet evaluated can be choosing more random ballots, one by one, until all preference relations in the TBRO have been established, or until all ballots have been used. If there is still an unevaluated pair (which requires that absolutely none of the voters express a preference between the pair), the order can be determined randomly. Now, if two candidates are in a beatpath tie with one another, the ordering of the candidates can be resolved in favor of the candidate who is more highly ranked in the TBRO.

 

Passed criteria

            Here are a few informal proofs which show that weighted pairwise meets certain well-known mathematical criteria which are used to evaluate voting methods with respect to sets of ordinal preference rankings. These proofs are almost entirely proofs that beatpath passes the criteria; there is rarely and cause for the cardinal aspect to enter into the analysis.

 

1. Mutual majority criterion: If there is a single majority of voters who rank every candidate in a set S₁ over every candidate outside S₁, then the winner should certainly be a member of S₁.

            Let’s say that there is a majority who prefer the members of S to all members outside S. Every S-insider will pairwise-beat every S-outsider, and hence there will be beatpaths from every S-insider to every S-outsider. Conversely, no S-outsider will pairwise-beat any S-insider, and there will be no beatpaths from any S-outsider to any S-insider. Hence only members of S can possibly win.

 

2. Condorcet efficiency: If there is one candidate who wins all of their pairwise comparisons, then this candidate should certainly win.

            If candidate A is a Condorcet winner, then there must be a beatpath from A to every other candidate, and there cannot be any beatpaths from any other candidate to A. Therefore A will certainly win.

 

3. Schwartz efficiency: An undominated set is a set of candidates such that no candidate within the set is pairwise beaten by any candidate outside the set. A minimal undominated set is an undominated set which doesn’t contain a smaller undominated set. The union of minimal undominated sets is called the GOCHA set (for Generalized Optimal-CHoice Axiom).[3] Schwartz efficiency requires that the winning candidate should always be a member of the GOCHA set.

            Let’s say that there is a candidate B who is not a member of a minimal undominated set. There must be some candidate A who is a member of a minimal undominated set, such that A has a beatpath to B. (Otherwise, B would have to be in a minimal undominated set.) B cannot have a beatpath against A, because B is not in A’s minimal undominated set. Hence B cannot win.

 

4. Pareto: If at least one voter ranks candidate A over candidate B, and zero voters rank candidate B over candidate A, then candidate B should certainly not win.

            Of course, A pairwise-beats B. Because no ballots rank A over B, we know that for any candidate X, if B pairwise-beats X, then A also pairwise-beats X, and that the strength of the B>X defeat cannot be greater than the A>X defeat. Also, for any candidate Y, if Y pairwise-beats A, then Y also pairwise-beats B, and the strength of the Y>A defeat cannot be greater than the Y>B defeat. Therefore, there can never be a beatpath from B to A that is stronger than the path from A to B. Hence, the best B can hope against A is a beatpath tie. However, if there is a beatpath tie, then the TBRO will certainly resolve the tie in favor of A, and therefore B cannot win.

 

5. Monotonicity: If candidate A wins with certainty according to a set of ballots, and some of the ballots are subsequently changed only in that A is ranked and/or rated higher on those ballots, then A should still win with certainty.

            If A’s rating and/or ranking is increased, it can only increase the strength of pairwise comparisons which A wins, and decrease the strength of pairwise comparisons which A loses. This cannot lead to an increase in any beatpath from any B to A, because a pairwise comparison in which A wins (the only comparisons that have potentially increased) will not be included in any beatpath from B to A. It also cannot lead to a decrease in any beatpath from A to B, because a pairwise comparison in which A loses (the only comparisons that have potentially decreased) will not be included in any beatpath from A to B. Hence, if A had a beatpath win against, B, A will still have a beatpath win over B.

 

6. Resolvability: This follows in a fairly straightforward way from the fact that beatpath is resolvable.[4] Basically, there is nothing about the cardinal pairwise principle which would make a beatpath result unresolvable.

 

Failed criteria

            Cardinal pairwise does not satisfy criteria which are incompatible with Condorcet efficiency, such as participation[5], consistency[6], later-no-harm[7], and later-no-help[8].

 

Also, cardinal pairwise does not satisfy independence of clones, unless the definition is adapted to require clones to be given the same ratings as one another, while still allowing for them to receive different rankings. For example,

40: R 100 > S 50 > T 40 > A 30 > B 0

25: A 100 > B 70 > R 60 > S 10 > T 0

35: B 100 > R 90 > S 20 > T 10 > A 0

            R, S, and T form a clone set, according to the standard definition. If R is included in the tally, R will win. If R is not included in the tally (but all of the other candidates are included), A will win.

 

            The maximizing provision causes the method to fail monotonicity. For example,

1: A 100 > B 60 > C 0 > D 0

1: C 100 > D 100 > A 50 > B 0

1: B 100 > C 100 > A 90 > D 0

1: D 100 > A 100 > B 50 > C 0

1: D 100 > C 100 > A 50 > B 0

1: B 100 > C 50 > D 0 > A 0

1: B 100 > D 100 > A 50 > C 0

1: C 100 > A 50 > B 0 > D 0

            A wins. But if you change 1: B 100 > C 50 > D 0 > A 0 into 1: B 100 > C 50 > A 0 > D 0, B wins. The reason is that D is no longer included in the Schwartz set, and the ratings on the ballot marked B 100, C 100, A 90, D 0 are subsequently maximized in scale to read B 100, C 100, A 0, thus strengthening the defeat against candidate A.