ComparisonsOne of the strongest reasons to prefer rated Condorcet methods to other Condorcet methods is that they can do better than other Condorcet methods in situations where cycles are strategically induced because they allow voters to prioritize which pairwise matchups they want to win. In the below example, the top line of voters honestly voted A>C>B>D, making C the Condorcet winner. But...:
Now A's 6 supporters naturally are not happy about that. What can they do to make their man win? There is one, and basically only one, change to their "A>C>B>D" vote they can make which causes A to win. That is to "bury" C with "A>B>D>C." (Well, the alternate burials A>D>B>C and A>D>C>B also work in this example, but they both are at-least-as-dishonest votes and anyway still count as burying C.) If they do this, we getThis problem is averted with Smith//Score or Smith//Approval if the C>A voters (voters who prefer C to A) move their approval threshold between C and A, because they can make C have 11 approvals to A's 10. Essentially, they can re-simulate the pairwise matchup between C and A (where C has 11 votes to A's 10) using min-max strategy to fix the result. This isn't as easy with Condorcet-IRV hybrid methods or defeat-dropping Condorcet methods; for the most part (ignoring things like the Tied at the top rule, etc.), the only way for C>A voters to fix the result is to Favorite Betray. In some sense, this all takes advantage of how rated methods have Nash Equilibriums on the Condorcet winner. Another thing to note about these methods' resistance to Favorite Betrayal is that even if voters do decide to Favorite Betray, they can still give a high rating to their favorite candidate; this means that even if, in a given election, a candidate fails to win due to Favorite Betrayal, it is still easy for voters to determine approximately how much support that candidate has (in addition, of course, to observing how many votes the candidate gets in head-to-head matchups against weak candidates, which gives a rough measure of how many voters support that candidate at all). Therefore, they can figure out whether they don't have to Favorite Betray in the next election.
whereupon A wins using every one of these Condorcet methods: Tideman ranked pairs, Basic Condorcet, Simpson-Kramer min-max, and Schulze beatpaths. (Success!)
#voters Their vote 6 A>B>D>C 2 B>C>A>D 3 B>A>C>D 2 C>D>A>B 2 C>B>A>D 5 D>C>A>B 1 D>A>C>B
One way to demonstrate the result in a rated Condorcet method election is to organize the candidates by Smith set ranking (if using a Smith-efficient hybrid) and then within each Smith set, organize the candidates by number of points/approvals, showing this number in the cell comparing each candidate to themselves. Also, a candidate's quality can be indicated simply by referring to which Smith set they were in and how many points they got; for example, with Smith//Approval, a candidate in the 5th Smith set (the Smith set when candidates in the Smith set are removed, with the new candidates in the Smith set forming the 2nd Smith set, etc.) with 54 approvals would be considered worse than anyone in the 1st through 4th Smith sets, as well as anyone in the 5th Smith set with more than 54 approvals. However, note that these methods are best used to indicate which candidate came closest to being the winner when all candidates remain in the election; the common "repeatedly remove the winner and recalculate the election result" method of finding the order of finish can yield a different result. 3-candidate example:
25 A| >B>C
34 B| >C>A
6 B>C| >A
35 C>A| >B
The Smith//Approval ranking by the method mentioned in the article is A>C>B, because all 3 candidates are in a cycle and the approvals are A 60 C 41 B 40. But if A drops out of the election, then the Smith//Approval ranking by either method becomes B>C, because B pairwise beats C, and thus becomes the only member of the 1st Smith set, with C being in the 2nd Smith set. So the "remove the winner and recalculate the result" method of finding the order of finish would say that the Smith//Approval ranking should be A>B>C.
Rated Condorcet methods have a stronger resistance to center squeeze than Category:Condorcet-IRV hybrid methods, because even when an artificial/strategic cycle is created, voters can rate the center candidate highly to make them win, rather than have the center candidate automatically eliminated unless they Favorite Betray.
Point of comparison between rated Condorcet methods and regular rated methods: if a mutual majority wish to make one of their preferred candidates win in a rated method, they must show maximal support for all of their candidates, not showing any distinction in preference between any of their candidates, and no support to any other candidates. By contrast, in a rated Condorcet method, because many of them pass the Smith criterion, the majority need only vote honestly to make their preferred candidates win, and can also show their preferences among all candidates. This further means that a minority that wants to maximally push for its preferred candidates can do so while showing preference among the majority's candidates without risking its own chances of winning as much.
Condorcet+cardinal methods that use rated ballots can sometimes benefit from the addition of an approval threshold. This is because if the score itself is used to determine both ranking and cardinal support, then a voter wishing to rank three candidates consecutively will have to give at least some support to their 2nd choice candidate. Using an approval threshold allows the voter to use the scores to rank the candidates and the approvals to express their rated support only for those candidates they want.
See rated pairwise preference ballot for ways to use cardinal information to decide the strength of each pairwise matchup.