Talk:Condorcet ranking

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If there are pairwise ties or cycles, do Condorcet rankings reflect this? For example, if there is a 3-candidate Smith Set and a 4th candidate, are the Smith candidates considered to be in a tie for 1st, with the 4th candidate in 2nd place? BetterVotingAdvocacy (talk) 22:35, 20 February 2020 (UTC)

I don't think it reflects cycles, though it could reflect ties. Reflecting cycles would be a "Smith ranking", not a Condorcet ranking. So a Condorcet ranking is only defined when there's a sequence of candidates so that A beats everybody, B beats everybody but A, C beats everybody but A and B and so on. A bit like criteria, that a method that always returns a Condorcet ranking doesn't say anything about what happens when there isn't one. Kristomun (talk) 00:25, 21 February 2020 (UTC)
I'd like to build a consensus for a "generalized Condorcet ranking criterion"; a voting method complies with this criterion if it (Much later edit: Ignore this small part) ~~(Edit: here's a more succinct way to put it: if we take any two groups of candidates, with the two groups not necessarily amounting to all candidates together, and any candidate in the first group can pairwise beat any candidate in the second group, then all candidates in the first group must be ranked higher than all candidates in the second group.)~~ (Edit 2: This is actually wrong, since it implies that in a 3-way Condorcet cycle, one candidate must be ranked above another above another above the first, which is impossible.) always says that A is at least as good as B when A pairwise beats or ties B. In essence, if a group of candidates pairwise beat all others, they must be ranked higher than all of them, but if within the group there is a cycle or pairwise ties, then the voting method can either rank some candidates in the group above others or show ties between some of them. From there, we can consider a "generalized Condorcet ranking" one where candidates in a cycle or pairwise ties are considered to be tied in the generalized ranking. The reason I'm proposing this is that it will help us pick out the "good" Smith-efficient methods (since I'd say this is at least one pretty decent standard of deciding what a good order of finish looks like), and further, helps us explain to the public how it is that Smith-efficient Condorcet methods in general go about deciding which candidates are better than others in the order of finish. In general, I just like the idea of clearly showing to people which groups of candidates are better than others based on pairwise preference, rather than solely pairwise preferences between two individual candidates. What do you think?
Edit 3: I think I have the definition down this time, and you're probably right that it should be called a "Smith ranking": take the definition of a Condorcet ranking, but replace the words "Condorcet winner" with Smith set, and make it so that if there is a multi-member Smith set, they need not all be ranked at the same place (1st, 2nd, etc.), so long as they are all ranked higher than the alternatives they dominate. Also, regarding Condorcet losers, I think I have a new analagous concept for that: the "Smith loser set", the smallest group of candidates such that all candidates in the group pairwise lose to all candidates not in the group. I'm discussing that at as well, but if it is a sensible concept, then it could replace the words "Condorcet loser" in the definition as well.
Edit 4: With regards to the "Smith loser set" idea, I think it's worth mentioning that it is not so simple to extend the idea in the same ways that the Smith set has been extended. For example, the Smith criterion requires that one of the Smith set candidates must win; if we tried to do this with the Smith loser set ("nobody in the Smith loser set should win") then we can end up with all candidates being eliminated; simple example is 1 voter voting A and another voting B. So it seems that a "Smith loser criterion" may have to be something along the lines of "candidates who are in the Smith loser set but not in the Smith set must not win." Also, I'm not sure of what use it might be, but a property analagous to ISDA for the Smith loser set might be "if a candidate is in the Smith loser set but not in the Smith set, then eliminating that candidate shouldn't change the result of the election." ISDA seems to imply this property automatically, and pretty much the only way I can imagine a voting method passing ISDA but not this property is if it is specifically designed to focus on Smith loser set candidates. The only remaining property that it seems would be interesting to extend would be the majority loser criterion to the mutual majority case (maybe "if a majority of voters prefer every other candidate over a group of candidates, nobody in that group of candidates should win." ?) BetterVotingAdvocacy (talk) 00:55, 21 February 2020 (UTC)
Your independence of Smith loser criterion (ISLDA) is the same as ISDA. Suppose you have an outcome with A=B>C=D>E=F being the Smith ranking. Then ISLDA says if you eliminate E and F, then the outcome will be the same. After doing so, your Smith ranking is A=B>C=D. So ISLDA says that if you eliminate C and D, then the outcome will be the same. Finally, your Smith ranking is A=B. But that's just the Smith set. Thus ISLDA is equivalent to ISDA.
Many other sets can be turned into rankings. For instance, the uncovered set could be turned into a ranking where the top-ranked candidates are the uncovered ones, then the next rank down consists of the candidates only covered by those that are top-ranked, and so on. Some people argue that Copeland's method is actually a set ranking due to how often it produces a tie. Kristomun (talk) 10:31, 21 February 2020 (UTC)
I think I can prove that when the Smith set and Smith loser set have any overlap, it is because they both include all candidates. The reasoning is that if we take the candidates that overlap between the two sets, logically they must simultaneously be defeated by everyone else in the Smith set but also beat everyone else in the Smith loser set; yet if all of these candidates are beaten by everyone in the Smith set, then they can't be in the Smith set, and similar reasoning goes for the Smith loser set. So if there is overlap, it is only possible when the Smith set and Smith lower sets are empty when excluding all of their overlapping candidates, because that way nobody pairwise beats the overlapping candidates and the overlapping candidates lose to nobody. So from here, we can simplify criterion-related extensions of the Smith loser set to simply say "voting methods shouldn't elect anyone in the Smith loser set unless this disqualifies everyone." BetterVotingAdvocacy (talk) 21:31, 21 February 2020 (UTC)
Well yes. Consider the Smith ranking as a flight of stairs. First step: smallest set of candidates who beat everybody else. Second step: smallest set of candidates who beat everybody else but the first step. Third step: smallest set of candidates who beat everybody else but the first two steps. The only way you can have only one step is if the cycle involves everybody, and the Smith loser set is simply the bottom-most step.
You could define a Smith loser criterion, but I think the reason nobody has done so already is that every method that passes it also tends to pass Smith unless the method is contrived to specifically pass the criterion. If you're going to stitch together two methods (e.g. Smith,IRV), there's no reason not to go the full mile and use the actual Smith set for your stitching and get Smith compliance essentially for free; and Condorcet methods that have only one type of logic, like Ranked Pairs or Schulze or Raynaud, usually either pass Smith directly or doesn't care about Smith and so also fails Smith loser. Kristomun (talk) 00:49, 22 February 2020 (UTC)
Does the Copeland ranking count as a Smith ranking? Given that Copeland is Smith-efficient, this seems to be the case, and if so, it could be a good idea to suggest that the simplest, most neutral way to produce a Smith ranking for a set of candidates is to use the Copeland ranking. Edit: In case it helps for a proof, according to this PDF ( the "Copeland set" (the candidate(s) with the uniquely highest Copeland score) is always in the Smith set. Edit 2: The answer is yes. Because of what I mentioned in my first edit, the solution to this question can be found using a generalized inquiry: if we remove some members of the Smith set, will the remaining members of the Smith set always be ranked higher in a Smith ranking than all other candidates? And the answer to that is "yes", since the lowest-Smith-ranked candidate of the Smith set will still pairwise beat all non-Smith candidates, and can only be pairwise beaten or tied by remaining members of the Smith set, therefore a Smith ranking must rank that lowest-Smith-ranked candidate and any candidates ranked higher than them strictly higher than all other candidates. BetterVotingAdvocacy (talk) 19:49, 24 February 2020 (UTC)

I think it'd be a good idea to have some kind of web service that can output a generalized Smith ranking of the candidates; this way, you can sidestep the argument over which Smith-efficient method to use and at least just see which groups of candidates are pairwise better than others. Does anybody know if someone like Eric Gorr would be willing to provide such a service? BetterVotingAdvocacy (talk) 00:15, 23 February 2020 (UTC)