Re: Condorcet Voting and Supermajorities (Re: [CONSTITUTIONAL AMENDMENT] Disambiguation of 4.1.5)
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
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