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Condorcet and FBC
184.108.40.206 changed the article from saying that most or all Condorcet methods fail FBC to claiming that all Condorcet methods fail. Is it proven that all Condorcet methods fail FBC?
- That was me (220.127.116.11). On the EM list I recently was able to show that a method that satisfies Condorcet necessarily has situations in which changing an equal ranking A=B to a strict ranking A>B on some ballots increases the probability that the winner is either A or B. This incentive isn't compatible with FBC. Kevin Venzke 07:02, 7 Jul 2005 (PDT)
- Why is that necessarily incompatible with FBC? Obviously, if A>B gives a probability that the winner is either A or B that is not only greater that what A=B gives but also greater than what B>A gives, that is a FBC failure. However, if both A>B and B>A give a greater probability that the winner is either A or B than what A=B gives, and A>B and B>A give the same probability as each other, then that is not necessarily a FBC failure. (As the criterion is currently stated, ranking one sincere co-favorite over another does not qualify as a favorite betrayal.) - DPJ, 2006-07-24 07:18 UTC
Criterion for lowest support
A better IRV needs a better example.
I have an improved version of IRV that does not fail in the example given. There is no need for any of the 10 supporters of A>B>C>D to change their votes. I have an example (#11) on my spreadsheet demonstrator that turns a candidate’s loss into a three way tie that requires a random draw to win. That’s less rewarding than the outright win granted in IRV. Therefore, I would like to claim that although it can still fail, it is a better version of IRV. (Is it still considered IRV? See the link on my user page). RalphInOttawa (talk) 09:09, 13 December 2023 (UTC)