I personally like the idea of using Bucklin to break a Condorcet cycle. Suppose alternatives Ax, Ay, and Az prevent there being a Condorcet winner. Then find the best ranks Rx, Ry, Rz for which a majority of the voters rank Ax, Ay, Az AT LEAST Rx,y,z respectively. If one of Rx, Ry, Rz is better than the others, that determines the winner. If Rx=Ry=Rz (or the two best are equal) then use the size of the respective majority (i.e. if 100 voters and 52 give Rx but 51 give Ry then Ax wins the tie with Ay.)
If both the R and #votes providing the majority that determines R are the same we have a true tie. Either flip a coin or have a runoff. -----Original Message----- From: election-methods-boun...@lists.electorama.com [mailto:election-methods-boun...@lists.electorama.com] On Behalf Of Terry Bouricius Sent: Thursday, May 14, 2009 2:48 PM To: Warren Smith; election-methods Subject: Re: [EM] Cramster question Warren, However, using first-choice plurality to settle Condorcet cycles could easily elect the Condorcet-loser (the candidate who loses in every pairwise match-up). There are many far superior cycle breakers. I personally favor ranked-pairs because it is both reasonable, and relatively easy to explain to lay people (unlike many cycle breakers). Terry Bouricius ---- Election-Methods mailing list - see http://electorama.com/em for list info