# 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

```