Re: Norman Petry and I (Ossipoff) recommended CSSD, but Schwartz Woodall is a better voting system for Debian

```On Thu, May 09, 2013 at 02:10:28PM +0100, Ian Jackson wrote:
> Michael Ossipoff writes ("Norman Petry and I (Ossipoff) recommended CSSD, but Schwartz Woodall is a better voting system for Debian"):
> > Example 1:
> >
> > Sincere preferences:
> >
> > 99: A>B>>C
> > 2: B>A>>C
> > 100: C>>(A=B)
> >
> > The A voters rank sincerely, and the B voters defect:
> >
> > 99: A>B
> > 2: B
> > 3: C
>
> In Debian's system, this will result in A winning.

Note that he corrected his example to 100: C

> A vs B:
>   99x "A>B"  count as preferring A to B
>    2x "B"    count as preferring B to A
>    3x "C"    count as preferring neither A to B, nor B to A

You are of course correct in this, for some reason I said B didn't
compare to anything, but it would get the default value so that
it's ranked lower than any voted value, and what it really says
is:
99x: (A>B)>C
2x: (B)>(A=C)
100x: (C)>(A=B)

>   So A defeats B.
>
> A vs C:
>   99x "A>B"  count as preferring A to C
>    2x "B"    count as preferring neither A to C, nor C to A
>    3x "C"    count as preferring C to A
>   So A defeats C.

So the corrected example would have 99 prefered A to C,en 100 C to
A, making C the winner, so 1 more prefering C over A than A over
C and making C defeat A.

> The Schwartz set contains only A.

The corrected example would contain C.

> > ----------------------------
> >
> > Here's another example in which the 3 factions are nearly equal in size:
> >
> > Sincere preferences:
> >
> > 33: A>B>>C
> > 32: B>A>>C
> > 34: C>>(A=B)
> >
> > Actual votes, when A voters co-operate and B voters defect:
> >
> > 33: A>B
> > 32: B
> > 34: C

So that needs to be translated to:
33: A>B>C
32: B>(A=C)
34: C>(A=B)

> A vs B:
>   33x "A>B"  count as A>B
>   32x "B"    count as B>A
>   34x "C"    count neither way
>   A defeats B by 33:32
>
> A vs C:
>   33x "A>B"  count as A>C
>   32x "B"    count neither way
>   34x "C"    count as C>A
>   C defeats A by 34:33
>
> B vs C:
>   33x "A>B"  count as B>C
>   32x "B"    count as B>C
>   34x "C"    count as C>B
>   B defeats C by 65:34
>
> The Schwarz set contains A,B,C.  We drop the weakest defeat, which is
> A>B by one vote because 33<34.

I think we shoud drop both both B and A because they are both
defeated by 1.  And C would be the winner.

Note that A defeats B, which defeats C, which defeats A again.

Kurt

```