[EM] More easily hand-counting three-slot Condorcet
Kevin Venzke
stepjak at yahoo.fr
Fri Nov 5 22:54:51 PST 2004
Jobst,
Thanks for taking the time to look at this idea.
--- Jobst Heitzig <heitzig-j at web.de> a écrit :
> But the vector
> (0,1,1,0,0,1,0,0,0)
> is linearly independent from
> (1,1,1,1,1,1,1,1,1), [...]
Interesting, this is precisely what I ended up doing to convince
myself that the magnitudes couldn't be pulled out. I'm surprised
to see it reproduced so readily.
> > I believe all rules are then equivalent to "Elect the least-
> > approved candidate who pairwise beats every candidate with
> > greater approval."
>
> That's true, but it would erroneously count those having A in slot 2 and
> B in slot 1 towards the strength of defeat A>B.
Hmm. I note that if A is going to be elected despite being defeated
pairwise by B, this must mean that A has greater approval than B, and
that A's wins will be locked first.
(The scenario I have in mind is:
Approval ranking: A>B>C. Pairwise: B>A>C>B. B wins.
If B's approval is simply reduced somehow, with no other changes:
Approval ranking: A>C>B. Now A wins.)
I think my suggestion is very similar to yours in spirit. They both
involve compressing the top two ranks.
> In accordance with my
> recent "grand compromise" proposal, I suggest to use instead the number
> of voters who have A in slot 1 or 2 and B in slot 3, which corresponds
> to V13+V23.
>
> Although also this number cannot be determined from the numbers you
> suggest to tally, it is not too expensive to make it possible by also
> counting all voters who have both A and B in slot 1 or both A and B in
> slot 2. This corresponds to counting V11+V22 also.
This would require a second matrix, wouldn't it? That would seem to
defeat the purpose somewhat!
Also, if you consider my second example:
49 B>>GN
24 G>>BN
27 N>G>B
I notice that the N>G win has a strength of 0, which the N>G>B voters
can adjust in a straightforward manner (by downranking G). I can't
put my finger on it, but it seems to me that the best placement of G
is less uncertain using my suggested measure of defeat strength.
(That is, I think the desire to register a G>B vote would far outweigh
the concern that if G beats N, the N>G>B vote strengthens that
victory. If I'm right about that, then I see it as a positive thing,
because we can get the most information from the N>G>B voters.)
I'm not yet able to picture clearly how your measure of defeat strength
would change everything, though.
Kevin Venzke
Vous manquez d’espace pour stocker vos mails ?
Yahoo! Mail vous offre GRATUITEMENT 100 Mo !
Créez votre Yahoo! Mail sur http://fr.benefits.yahoo.com/
Le nouveau Yahoo! Messenger est arrivé ! Découvrez toutes les nouveautés pour dialoguer instantanément avec vos amis. A télécharger gratuitement sur http://fr.messenger.yahoo.com
More information about the Election-Methods
mailing list