Anonymous user
Talk:Favorite betrayal criterion: Difference between revisions
Added query on incompatibility of Condercet and FBC
imported>KVenzke No edit summary |
(Added query on incompatibility of Condercet and FBC) |
||
Line 3:
That was me (12.73.128.232). 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. [[User:KVenzke|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
| |||