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
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