On Tue, Dec 10, 2002 at 02:50:11PM -0500, Raul Miller wrote: > [b] Gives a result which is less like condorcet than "drop all failed > supermajority before CpSSD". Uh, how do you figure: 40 A B D (A requires 3:1 supermajority, D is the default option) 20 D B A which Condorcet would treats as "A wins, B comes second, D comes last", is treated "more like Condorcet" by: D defeats A (60:40) (scaled 3:1) A defeats B (40:20) B defeats D (40:20) Drop weakest defeats, leaving D defeats A; Hence B and D draw ? If you want a more complicated example, try: 40 A B C F (A requires 3:1 supermajority, F is the default option) 10 C B F A 10 F C B A which Condorcet would rank as A first, B second, C third and F last; but under Hybrid would be treated as: F defeats A (60:40) (scaled 3:1 from 20:40) C defeats F (50:10) B defeats F (50:10) A defeats B (40:20) <--- every defeat below this one must be dropped B defeats C (40:20) A defeats C (40:20) with the three weakest defeats eliminated, leaving F defeats A, C defeats F, B defeats F, and a draw between B and C. > I don't think such cases exist (I'm working on how to show this). On the > other hand, there are cases where "drop all failed supermajority before > CpSSD" gives results which are less like condorcet than "Hybrid Theory". (Ah, assertions without examples. How helpful.) Define "like Condorcet". I think the simplest definition, and one that makes sense, is to say "Gives the same result as Condorcet would if you only consider the options that passed supermajority and quorum", which "drop all failed supermajority options before CpSSD" obviously matches. Here's a simpler criterion: given two options both of which pass their supermajority requirement and quorum, and where there's a beatpath from A to B, but no beat path from B to A (that is, A transitively defeats B in the raw, unscaled votes), B cannot win the election (the rationale being that a majority prefer A, and there aren't any cycles to confuse the issue). Certainly "drop all failed supermajority options, then do CpSSD" loses information; but the information it loses is essentially irrelevant. It's not even comparable to the normal case of ties within the Schwartz set: in that case you have a real, sincere cycle amongst the preferences in the Schwartz set: if you have a cycle A beats B beats C beats A, then *each* of A, B and C are in the Schwartz set; whereas there's simply no sincere cycle with just "B and D" and "B and C" in the above examples: B is clearly preferred. 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. ``If you don't do it now, you'll be one year older when you do.''
Attachment:
pgp6tTSK_u6DJ.pgp
Description: PGP signature