On Thu, 6 Mar 2003, Forest Simmons wrote: > > (8) We could combine your method (2) with method (6), by finding the max > of each row in your (transpose of pairwise) matrix and then subtracting > the min of the corresponding column. The candidate corresponding to the > minimum such difference would be the winner. > > [Note that this is similar to MinMax (margins) but with the subtraction > done after the row and column max and min operations.] > > This combination method (8) would take care of the anomalous results at > the end of your message. > > Note for example, that disaster associated with > > 1000: A, 1000: B, 1: AC, and 1: BC > > is avoided by taking into account the offensive prowess of candidates A > and B in comparison to C who is relatively strong defensively but weak > offensively. > > In general method (8) doesn't always pick the beats all winner. I don't > know if it does when restricted to ballots of resolution two (i.e. > approval ballots). I suspect there may be some cases where it doesn't > pick the approval/beats-all winner (which is not necessarily a defect in > this context).
In fact, your first example still works under this modification: Your respective row maxima were 15, 19, 22, and 27. Your respective row minima were 10, 12, 15, and 9. The respective differences are 5, 7, 7, and 18. So A is the winner by the modified median method! Also, I believe that your median method and the modified median method both satisfy the FBC. Forest _______________________________________________ Election-methods mailing list [EMAIL PROTECTED] http://lists.electorama.com/listinfo.cgi/election-methods-electorama.com