Ranked Pairs: Difference between revisions
Moved some of the Notes to an advantages and disadvantages section, and added Munger paper advocating for RP.
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We assume 48% of voters vote A>B>C, 3% votes B>C>A, 49% votes C>A>B. A>B in 97% (locked), C>A in 52% (locked) and B>C in 51% (not locked since there is a cycle). Thus, we have C>A>B and C is the winner. However, if we make a [[Borda count]] (2 points for first place 1 point for second place) A has 145 points and C has only 101 points and thus one could think that A deserves more to win.
== Advantages and disadvantages ==
Ranked Pairs is [[Smith-efficient]], because no Smith set member can be beaten by a candidate not in the Smith set, and therefore any candidate not in the Smith set can't have their defeats to Smith set members discarded during the RP procedure, so they can't become the Condorcet winner. ▼
Ranked Pairs passes the [[Independence of Smith-dominated Alternatives]] criterion, because the only cycles for RP to potentially resolve will always be between Smith set members. Because of this, all candidates not in the Smith set can be eliminated before starting the procedure, reducing the number of operations needed to be done to find the winner. In addition, Ranked Pairs, like [[Schulze]], is equivalent to [[Minimax]] when there are 3 or fewer candidates with no pairwise ties between them, so if the Smith set has 3 or fewer candidates in it with no pairwise ties between them, [[Smith//Minimax]] can be run instead to find/demonstrate the RP winner.▼
While Ranked Pairs behaves similarly to [[Schulze]], Ranked Pairs passes [[local independence of irrelevant alternatives]] whereas Schulze does not. Some authors argue that the Ranked Pairs method is more intuitive and easier to understand than Schulze as well.<ref name="Munger 2023 pp. 434–444">{{cite journal | last=Munger | first=Charles T. | title=The best Condorcet-compatible election method: Ranked Pairs | journal=Constitutional Political Economy | volume=34 | issue=3 | date=2023 | issn=1043-4062 | doi=10.1007/s10602-022-09382-w | pages=434–444}}</ref>
One disadvantage of Ranked Pairs is that there's no easy way to detect ties for first place: determining whether there exists a way to break ties between pairwise victories so that a given candidate wins is NP-complete.<ref name="Brill">{{cite journal | last=Brill | first=Markus | last2=Fischer | first2=Felix | title=The Price of Neutrality for the Ranked Pairs Method | journal=Proceedings of the AAAI Conference on Artificial Intelligence | publisher=Association for the Advancement of Artificial Intelligence (AAAI) | volume=26 | issue=1 | date=2012-07-26 | issn=2374-3468 | doi=10.1609/aaai.v26i1.8250 | pages=1299–1305}}</ref>. While it is possible to break pairwise ties fairly, doing so leads Ranked Pairs to declare one of the winners the sole winner, without giving information about whether other candidates could have won were the ties broken differently.▼
== Notes ==
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When ignoring struckthrough (non-locked in) pairwise victories, C is the only candidate with no pairwise defeats, and thus is the RP winner. The RP ranking is C>A>B, since C pairwise beats all others, A pairwise beats everyone except C, and B pairwise loses to everyone (when ignoring the defeats ignored by the RP procedure).
▲Ranked Pairs is [[Smith-efficient]], because no Smith set member can be beaten by a candidate not in the Smith set, and therefore any candidate not in the Smith set can't have their defeats to Smith set members discarded during the RP procedure, so they can't become the Condorcet winner.
▲Ranked Pairs passes the [[Independence of Smith-dominated Alternatives]] criterion, because the only cycles for RP to potentially resolve will always be between Smith set members. Because of this, all candidates not in the Smith set can be eliminated before starting the procedure, reducing the number of operations needed to be done to find the winner. In addition, Ranked Pairs, like [[Schulze]], is equivalent to [[Minimax]] when there are 3 or fewer candidates with no pairwise ties between them, so if the Smith set has 3 or fewer candidates in it with no pairwise ties between them, [[Smith//Minimax]] can be run instead to find/demonstrate the RP winner.
▲One disadvantage of Ranked Pairs is that there's no easy way to detect ties for first place: determining whether there exists a way to break ties between pairwise victories so that a given candidate wins is NP-complete.<ref name="Brill">{{cite journal | last=Brill | first=Markus | last2=Fischer | first2=Felix | title=The Price of Neutrality for the Ranked Pairs Method | journal=Proceedings of the AAAI Conference on Artificial Intelligence | publisher=Association for the Advancement of Artificial Intelligence (AAAI) | volume=26 | issue=1 | date=2012-07-26 | issn=2374-3468 | doi=10.1609/aaai.v26i1.8250 | pages=1299–1305}}</ref>. While it is possible to break pairwise ties fairly, doing so leads Ranked Pairs to declare one of the winners the sole winner, without giving information about whether other candidates could have won were the ties broken differently.
== References ==
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