Re: Condorcet Voting and Supermajorities (Re: [CONSTITUTIONAL AMENDMENT] Disambiguation of 4.1.5)
> On Fri, Dec 01, 2000 at 09:31:13AM -0500, Raul Miller wrote:
> > On Fri, Dec 01, 2000 at 06:54:32AM -0500, Raul Miller wrote:
> > > > > > > > However, with the 3:1 supermajority which affects A, you get:
> > > > > > > > 10 : 0 B:C
> > > > > > > > 3 1/3: 0 A:B
> > > > > > > > 3 1/3: 0 A:C
> > > > > > > > B wins.
On Sat, Dec 02, 2000 at 12:45:53AM +1000, Anthony Towns wrote:
> This summary is completely wrong for any Condorcet scheme.
>
> A dominates B (by a reduced margin of 3.3 to 0 rather than 10 to 0)
> A dominates C (also by a reduced margin)
> B dominates C (by its original margin)
>
> A thus wins by dominating all others.
>
> Whatever you're using to declare B the winner above, it's not a Condorcet
> method.
My hypothesis is: if everyone votes for all options, any smith set
which combines an option with a supermajority and an option which does
not, will result in a lose for the supermajority if you're using the
condorcet method.
> The rest of your message, and the conclusion that
> Condorced+Supermajority isn't possible is thus invalid.
It's a simple observation about the numbers involved.
However, I've been wrong before -- can you construct a counterexample
to my hypothesis?
Alternatively, can you show me how my hypothesis is irrelevant?
Thanks,
--
Raul
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