[EM] Cramster question

[Paul Kislanko]

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-bounces at lists.electorama.com
[mailto:election-methods-bounces at 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