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Re: Nov 19 draft of voting amendment

On Tue, Nov 19, 2002 at 11:58:35AM -0500, Raul Miller wrote:
> > It's probably worth comparing the strategies possible with this draft [...]

On Wed, Nov 20, 2002 at 05:06:01AM +1000, Anthony Towns wrote:
> I'm going to ignore the fact you meant wrt quorums not supermajorities.

*blush*  Thanks.

> Consider 100 voters, a constitutional amendment, A, and a set of
> conscientious objectors. The objectors vote:
> 	30 D A
> This should, surely, be enough to stop "A" winning, no? Isn't that the point
> of a supermajority requirement?
> If the remaining 70 people vote:
> 	70 A D
> they'll succeed. If, however, they can introduce a cycle amongst A, D and
> something else, and have D:A 90:70 be the weakest defeat, like so:
> Introducing the new option, B, no supermajority requirement.  They convince
> a bunch of the conscientious objectors to support it.
> 	20 B D A
> 	10 D A
> 	70 A B D
> A defeats B (100:0), B defeats D (90:10), D defeats A (90:70). "D beats A"
> is then the weakest defeat, and A wins, in spite of the 30 person bloc
> trying to ensure its defeat.
> I *believe* that 30 of 100 people voting "D" as their first option can
> successfully ensure that a supermajority requiring option they're opposed
> to will be defeated, but why should they be required to insincerely say
> they wouldn't be satisfied with B in order to exercise their prerogative
> as a superminority?

Because Condorcet allows weak transitive defeats to be overridden by
strong transitive defeats.  I thought you understood this.

[This is why I had been arguing for using supermajority scaling on
all options.]

> >     2. We drop the weakest defeats from the Schwartz set until there
> >        are no more defeats in the Schwartz set:
> >          a. An option A is in the Schwartz set if for all options B,
> >             either A transitively defeats B, or B does not transitively
> >             defeat A.
> >          h. A defeat is in the Schwartz set if both of its options are
> >             in the Schwartz set.
> >          i. A defeat (R,S) is dropped by making N(S,R) the same as N(R,S).
> >             Once a defeat is dropped it must stay dropped.
> If you have a Schwartz set {A,B,C} and eliminate a defeat, can you then end
> up putting some {D} into the Schwartz set?

In general: for every option which is not in the Schwartz set there
must also be a defeat of that option by some option which it does not
transitively defeat.

Eliminating defeats of options in the Schwartz set can not eliminate
defeats of options which are not in the Schwartz set.

In other words: no, eliminating defeats from the Schwartz set can't put
any options into the Schwartz set which weren't already there.

> If D is not in the Schwartz set
> then there's some option X which transitively defeats D, but D does not
> transitively defeat D.

It's not safe, in general, to say that D does not transitively defeat D.
For example, imagine that the ballot includes options E and F, and:
D defeats E, E defeats F and F defeats D.  Also imagine that A defeats D.


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