Pairwise counting: Difference between revisions
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Multi-winner methods that use pairwise counting, such as [[CPO-STV]] and [[Schulze STV]], instead of doing pairwise matchups between individual candidates, do pairwise matchups between sets of candidates (called winner sets).
One of the notable aspects of pairwise counting is that it can be used to find a Condorcet winner or member of the Smith set in a simple manner without needing to be done with written ballots; see [[:Category:Sequential comparison Condorcet methods]] for more information.
Writeup on solving the write-in issue for pairwise counting:<blockquote>So, the problem is that if one voter indicates they prefer (Write-in candidate)>A>B, and another voter prefers B>A, you now have to explicitly figure out how to show that the second voter prefers B and A over the write-in candidate before you even know who that candidate is.
A generalized solution is to, for each candidate the voter ranks or rated, give them one vote, and then for each candidate ranked or rated above that candidate on the voter's ballot, indicate a negative vote in that pairwise matchup. That is, the above B>A voter would be considered as voting for B and A, with -1 vote recorded for A>B. Then, for each vote that a candidate has, that is one vote they get in every head-to-head matchup against all other candidates. So the B>A voter gives 1 vote to B>write-in and A>write-in, and 1 vote to both B>A and A>B, but because of the -1 vote for A>B, this just becomes 1 vote for B>A.
For equally ranked candidates, such as A=B>C>D, you put a negative vote for both A>B and B>A. Bonus: The votes for each candidate can be placed in the blank cell comparing themselves to themselves in the pairwise matrix i.e. A>A would contain A's votes.<ref>{{Cite web|url=https://www.reddit.com/r/EndFPTP/comments/fsa4np/possible_solution_to_the_condorcet_writein_problem/|title=Possible solution to the Condorcet write-in problem|last=|first=|date=|website=|url-status=live|archive-url=|archive-date=|access-date=}}</ref></blockquote>{{reflist|group=nb}}
== References ==
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