Re: Nov 18 draft of vote counting methodology

[Raul Miller <moth@debian.org>]

The notation I used in my response to Clinton was wrong for the draft
I proposed yesterday and the draft I proposed today.  It reflects a
[more complex] draft I'd been thinking about, but not the draft I wrote.

I apologize for that.  Here's a better response to Clinton.

On Wed, Nov 20, 2002 at 01:11:43AM +1100, Clinton Mead wrote:
> > I would like to know the result of this ballot.
> > 
> > ACB
> > ACB
> > CBA
> > CBA
> > CBA
> > 
> > Where A is default and C requires a supermajority (as far as I can tell 
> > its A, even though B, not needing a supermajority, beat it pairwise).

I'm going to assume a supermajority ratio of 2:1

N(A,B) 2;  N(B,A) 3
N(A,C) 4;  N(C,A) 3
N(B,C) 0;  N(C,B) 5

(A,B) 2:3 not a defeat
(A,C) 4:3 A defeats C
(B,A) 3:2 B defeats A
(B,C) 0:5 not a defeat
(C,A) 3:4 not a defeat
(C,B) 5:0 C defeats B

All options transitively defeat all other options.

The weakest defeat is (B,A), eliminating it leaves:

N(A,B) 3;  N(B,A) 3
N(A,C) 4;  N(C,A) 3
N(B,C) 0;  N(C,B) 5

(A,B) 3:3 not a defeat
(A,C) 4:3 A defeats C
(B,A) 3:3 not a defeat
(B,C) 0:5 not a defeat
(C,A) 3:4 not a defeat
(C,B) 5:0 C defeats B

Now, A transitively defeats all other options and no other option
transitively defeats A.  A wins.  [No options prefer A over the default
option, because A is the default option, so even after step 3, A wins.]

> > As far as I can tell, C is the winner of step two, but defeated in step 
> > 3 by the default option supermajority veto.

There is no "default option supermajority veto" in step 3, in the
Nov 18 draft.

-- 
Raul