> On Fri, Dec 01, 2000 at 09:31:13AM -0500, Raul Miller wrote: > > On Fri, Dec 01, 2000 at 06:54:32AM -0500, Raul Miller wrote: > > > > > > > > However, with the 3:1 supermajority which affects A, you get: > > > > > > > > 10 : 0 B:C > > > > > > > > 3 1/3: 0 A:B > > > > > > > > 3 1/3: 0 A:C > > > > > > > > B wins.
On Sat, Dec 02, 2000 at 12:45:53AM +1000, Anthony Towns wrote: > This summary is completely wrong for any Condorcet scheme. > > A dominates B (by a reduced margin of 3.3 to 0 rather than 10 to 0) > A dominates C (also by a reduced margin) > B dominates C (by its original margin) > > A thus wins by dominating all others. > > Whatever you're using to declare B the winner above, it's not a Condorcet > method. My hypothesis is: if everyone votes for all options, any smith set which combines an option with a supermajority and an option which does not, will result in a lose for the supermajority if you're using the condorcet method. > The rest of your message, and the conclusion that > Condorced+Supermajority isn't possible is thus invalid. It's a simple observation about the numbers involved. However, I've been wrong before -- can you construct a counterexample to my hypothesis? Alternatively, can you show me how my hypothesis is irrelevant? Thanks, -- Raul
