[EM] Three rounds

[Kristofer Munsterhjelm]

Raph Frank wrote:
> On Fri, Nov 14, 2008 at 8:56 AM, Kristofer Munsterhjelm
> <km-elmet at broadpark.no> 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/election-methods-list@eskimo.com/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.