Talk:Favorite Betrayal criterion

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Jump to: navigation, search 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 ( 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