Jobst, Here's another direction to take the idea of D2MAC:
Voters rate options as 1, 2, or 3 according to whether they are good, better, or best, respectively. All other candidates are rated at zero by default. Six ballots B1, B2, ... B6, are drawn at random. If there is an option that is rated above zero on all six of these ballots, then (of these options) the option that is rated above zero on the most (other) ballots wins the election. Else if there is an option that is rated above 1 on the first three drawn ballots, then (of these options) the one rated above zero on the most (other) ballots wins the election. Else of the options rated "best" on ballot B1, the one rated above zero on the most ballots wins the election. If I am not mistaken, this variation of FAWRB should give the compromise A a better chance in our second test case: 33 A1>A>A2 33 A2>A>A1 33 B Would it hurt C's chances in our first test case? 50 A>C 50 B>C Forest ---- Election-Methods mailing list - see http://electorama.com/em for list info
