Juho wrote: > My usual example is this: > 25:A>B>D>C > 25:B>C>D>A > 25:C>A>D>B > 24:D > > There are four parties of about equal size. D is the Condorcet loser > but on the other hand she is two votes short of being the Condorcet > winner. The other three candidates are badly cyclic and would each > need 26 additional votes to become Condorcet winners. In this light D > is not a bad winner after all. Not electing the Condorcet loser is a > good property in 99.9% of the elections but in the remaining tricky > cases things may look different.
I ran the example through the Rob LeGrand's election calculator at http://cec.wustl.edu/~rhl1/rbvote/calc.html and got the following: Winner: (A,B,C, requires tiebreaker) Baldwin* Black* Borda* Copeland* Nanson* Raynaud* Schulze* Small* Tideman* Winnder: D Dodgson Simpson Which seemed like an odd result to me. Any idea why Dodgson and Simpson gave the Condorcet loser? On a tangential note, as a test I tried it with IFNOP to see if it would resolve there, and got the following: 25:A>B>D>C 25:B>C>D>A 25:C>A>D>B 24:D>A=B=C Left hand side is normal comparison (>), right hand side is reverse (<), number in parenthesis is the difference. 25 AB 25 (0) 25 AC 25 (0) 50 AD 49 (1) 25 BC 25 (0) 50 BD 49 (1) 50 CD 49 (1) Since D is the Condorcet loser, it is dropped. AB 0+0+0+25+0+0 AC 0+0+0+0+0+25 BA 0+0+0+0+0+25 BC 0+0+0+25+0+0 CA 0+0+0+25+0+0 CB 0+0+0+0+0+25 This gives A=B=C -- a tiebreaker would be needed with this method as well. (It's D's voters' fault for not giving a complete preference, and an election giving D the win would seem to open such methods for manipulation.) A runoff election between A, B, and C would probably be the fairest method. Michael Rouse [EMAIL PROTECTED] ---- election-methods mailing list - see http://electorama.com/em for list info
