On Thu, Nov 23, 2000 at 08:04:21AM -0500, Raul Miller wrote: > On Tue, Nov 21, 2000 at 07:42:44PM +1000, Anthony Towns wrote: > > > In really rare cases this might lead to paradoxical situations where the > > > winning option doesn't have the required approval rating, but a lesser > > > option does. > > Some possibilities: > > a) A clear condorcet winner, that doesn't have enough of a > > supermajority to succeed. > Supermajority basically means that yes votes have fractional > significance. You don't have a clear winner if you don't have > enough votes -- unless you pretend that the yes votes have some > different significance? Suppose you have the following ballot, and the following votes: A: Remove non-free B: Support non-free F: Further discussion 60 votes ABF (Would prefer to remove non-free, but either is okay) 50 votes BFA (Insist that non-free not be removed) then A dominates B and F, 60 to 50, and B dominates F 110 to 0. The clear condorcet winner is A since it dominates all the other options (ie, it's what the majority of voters prefer) yet it doesn't have enough votes to claim a 3:1 supermajority. > > b) A tie for first place (ie, the schwartz set has two or more > > options in it), where "further discussion" is one of the > > equal winners, and it pairwise beats whatever is chosen as > > the real winner. > If this is a true tie, we need a tie-breaking vote (casting vote). That > would mean it's up to the leader for the stuff we're talking about here. A tie for first place in condorcet terms is more complicated than that. For example: 30 ABC 25 BCA 35 CAB A dominates B (65 to 25), B dominates C (55 to 35) and C dominates A (60 to 30) to complete the cycle. A single casting vote isn't enough to reverse any of them: you'd need to change at least 10 "casting" votes, which could reverse the B dominates C pair and let C win by dominating all others. This is roughly how Tideman works, I think (ie, by choosing the option that'd require the fewest swaps to make it the clear winner). > > c) A tie for first place where all the winners beat further > > discussion, but the winner selected by whichever tie breaker's > > used requires a supermajority that it doesn't have, and one > > of the other winners has all the majority it needs (because > > it only requires a smaller one, say) > That wouldn't have been a tie. By "tie" I mean an example like the above. I'm fairly sure the above case is possible. Cheers, aj -- Anthony Towns <aj@humbug.org.au> <http://azure.humbug.org.au/~aj/> I don't speak for anyone save myself. GPG signed mail preferred. ``Thanks to all avid pokers out there'' -- linux.conf.au, 17-20 January 2001
Attachment:
pgpWGCYLECCg5.pgp
Description: PGP signature