Raph Frank wrote:
On Fri, Nov 14, 2008 at 8:56 AM, Kristofer Munsterhjelm
<[EMAIL PROTECTED]> wrote:
This does mean that a party can crowd out its competitors by running two
candidates of the exact same position. On the other hand, that may be what
you want, since one could reason that this brings a competition of quality
to the center position, where the two best centrists would be picked for the
runoff. That doesn't give the people much to discuss between the first and
second rounds, though, since the candidates' position would be identical.
Their positions would likely be similar but not identical, especially
in a multi dimensional political space.
The campaign would come down to questions of capability as a
representative and small policy differences.
One possible issue would be a small turnout at the second round. This
might encourage them to appeal to extremists.
This, in turn, may cause the runoff to have Range-like effects. Say that
the runoff is held with two centrists as the candidates. Then voters who
feels the same way about both candidates may not bother to show up, even
though they prefer one candidate to the other.
If that effect is too prevalent, it could be confused for an apathetic
populace.
First round, use a method like Schulze to get a
social ordering. Pick the first and second place candidates on that social
ordering for the second round.
Right, that is what I was thinking. With any condorcet method, you
could just say pick the winner and then pick the winner excluding the
first winner, but I think most condorcet completion methods generate a
complete ordering.
I mention a complete ordering since some methods may act differently if
you eliminate the winner (particularly if they're nonmonotonic). Using
the complete ordering seems more sensible in that case.
Another option would be to pick the 2 most approved candidates for the
2nd round.
Then you'd need an approval cutoff, or plain Approval. I think Approval
would satisfy IIA if you use a constant strategy (compare all candidates
to an objective standard and approve those better than that standard),
but since the best Approval strategies are relative, there may still be
a point in an Approval runoff.
That's what D'Hondt without lists does; or rather, it deweights those
preferences that are lower than the winner of the first round, since the
voters already "got what they wanted" on a higher preference. (Of course, I
would use Sainte-Laguë instead of D'Hondt, but that's an implementation
detail.)
Interesting.
The process is that you vote for your top choice that is still in the
running but it is deweighted by the number of higher choices who have
already been elected?
A vote of A>B>C would vote for C if A and B were elected at a weight
of 1/5 strength (assuming Sainte-Lague)?
How are eliminations handled?
To paraphrase from the post
(http://www.mail-archive.com/[EMAIL PROTECTED]/msg08230.html
), D'Hondt without lists has this rule:
Downweight any preferences where both candidates compared are ranked
below k elected candidates, by f(k).
For D'Hondt, f(x) is 1/(x+1), or [1; 1/2; 1/3, ...], counting from 0. In
the case of Sainte-Laguë, f(x) is 1/(2x + 1), or [1; 1/3; 1/5, ...].
So in your A>B>C case, B>C would have weight 1/3. If you had A>B>C>D,
then C>D would have weight 1/5.
Eliminations are handled by removing already elected candidates from the
ordering. For instance, if A's elected and the result for the second
round is A > B > C > D, A is removed to make B > C > D, and B is
elected. This limits IIA oddness.
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