On Thu, Nov 23, 2000 at 11:31:44PM +1000, Anthony Towns wrote: > 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.
Nope. With a 3:1 supermajority, the 60 raw votes for A are equivalent to 20 real votes, so A does not dominate B. Furthermore, you've not specified quorum -- F gets quorum votes, automatically. > > > 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). According to the debian constitution, in this circumstance you eliminate B and retally: 30 AC 60 CA C wins. [Look for "Transferrable Vote" in the constitution.] > > > 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. I guess the above covers it. -- Raul
