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
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