Mike, yes it is the same as the one you repeated at the end of your reply below.
But notice that in the ballot set 49 C 27 A>B 24 B there are two Acquiescing Majorities, namely both {A, B} and {B, C}, and that C has more top votes than B. Forest > From: MIKE OSSIPOFF > To: > Subject: [EM] Acquiescing majority MMT > Message-ID: > Content-Type: text/plain; charset="iso-8859-1" > > > Forest-- > > Let's find out what its properties are. > > Just preliminarily, it sounds like one of the MMT ideas that I > considered. > > But it seemed to me, at the time (if it's the same method I was > considering) that, if a ballot can be counted in that majority > merely by rating each candidate > in the set equal to or over every candidate outside the set, > then, in the ABE, the B votes could > rate A at bottom, with C, and still be part of the relevant > majority. So there there is > the set required by the acquiescing rule, and there is one > candidate rated above bottom > by everyone in that set: Candidate B. > > So, the method that I'd considered wouldn't pass in the ABE. I > don't know if the method > you describe is the same one, but, preliminarily, it sounds similar. > > But maybe not. Any possibility could yield improvement. > > My definition of that set was something like this: > > A set of candidates rated equal to or over everyone outside the > set by each member of the > same majority of the voters. > > Mike Ossipoff > > > -------------- next part -------------- > An HTML attachment was scrubbed... > URL: > electorama.com/attachments/20111209/a13f00db/attachment.htm> > ------------------------------ > > _______________________________________________ > Election-Methods mailing list > Election-Methods@lists.electorama.com > http://lists.electorama.com/listinfo.cgi/election-methods- > electorama.com > > End of Election-Methods Digest, Vol 90, Issue 24 > ************************************************ > ---- Election-Methods mailing list - see http://electorama.com/em for list info