Jobst Heitzig <[EMAIL PROTECTED]> >> This is not independent of cloning ... in a big way :). > >What makes you think so? Assuming we add a clone y of x so that all >voters have x and y neighboured in their ranking and so that the new >initial proportions of papers naming either x or y is the same as the >original proportion of papers naming x, it seems that each voter will >return exactly the same number of papers naming either x or y as they >returned papers naming x originally. This seems to indicate that the >procedure is clone-proof all right, isn't it?
Yeah, I misread the procedure. I missed the doubling the ballots step. If there are 10 candidates with 9 from party B and 1 from party A all voters return 10 papers not 5 as I thought. 'A' supporters return 4 A's and 6 from B 'B' supporters return 10 B's The new odds are 4/20 = 20%, this is better for A than the original lottery (10%) and presumably in the limit will end up 50/50 as required. >> If so, there is an odd number of winners in that case, which might >> prove problematic. >This is not obvious to me. Can you prove it? I guess I just assumed as all results I have ever seen are 3 or 5 in the tie. It might be true though :). Some simple examples elections: 2 candidates A>B B>A all candidate tie 3 candidates A>B>C B>C>A C>A>B result: A>B>C>A (circle) 4 candidates A>B>C>D B>C>D>A C>D>A>B D>A>B>C Result: total tie, removing 1 vote gives 3 way circular tie. I can't see an obvious way to get the 4 to tie. ___________________________________________________ Try the New Netscape Mail Today! Virtually Spam-Free | More Storage | Import Your Contact List http://mail.netscape.com ---- election-methods mailing list - see http://electorama.com/em for list info
