[EM] Throwing my hat into the ring, possibly to get trampled

[Nicholas Buckner]

Thank you for the article, as it was informative. It is very true that
Elimination methods tend to eliminate candidates who could go onto
become winners.

QLTD doesn't have a single loser to eliminate (it doesn't mention
losers much either). In fact I worn against first-past-the-post
methods (vanilla QLTD) (that's why I go with the "converse" way--the
elimination way), as they are susceptible to Burying mentioned here
http://en.wikipedia.org/wiki/Tactical_voting

what I think the original article
(http://www.mcdougall.org.uk/VM/ISSUE6/P4.HTM) tries to show is QLTD
has a problem with a subset criterion to monotocity he called
mono-add-top (which is very close to the Participation criterion).

My method doesn't have that problem.  Let me use his example (though I
have some mild problems with the Droop Quota now, I'll still use it
for these calculations).

Election 2:
abcdef     12 votes
cabdef     11 votes
bcadef     10 votes
def        27 votes

when adding: ad 6 votes

Pre add:
Votes: 60
Quota: 31

Round 1:
1: 2.8
2: 2.818182
3: 2.833333
4: 3.121212
5: 4.121212
6: 5.121212
Candidate 6 has the worst multiplier and was removed.

Round 2:
1: 2.421053
2: 2.450000
3: 2.476190
4: 3.121212
5: 4.121212
Candidate 5 has the worst multiplier and was removed.

Round 3:
1: 1.950000
2: 2.000000
3: 2.047619
4: 3.121212
Candidate 4 has the worst multiplier and was removed.

Round 4:
1: 1.500000
2: 1.571429
3: 1.578947
Candidate 3 has the worst multiplier and was removed.

Round 5:
1: 0.849315
2: 1.205479
Candidate 2 has the worst multiplier and was removed.

Final order from worst to best: 6, 5, 4, 3, 2, 1.

Post add:
Votes: 66
Quota: 34

Round 1:
1: 2.500000
2: 2.960000
3: 2.962936
4: 3.030303
5: 4.115942
6: 5.072464
Candidate 6 has the worst multiplier and was removed.

Round 2:
1: 2.263158
2: 2.545455
3: 2.565217
4: 3.030303
5: 4.085714

Candidate 5 has the worst multiplier and was removed.

Round 3:
1: 1.800000
2: 2.130436
3: 2.166667
4: 3.030303

Candidate 4 has the worst multiplier and was removed.

Round 4:
1: 1.350000
2: 1.625000
3: 1.636364

Candidate 3 has the worst multiplier and was removed.

Round 5:
1: 0.800000
2: 1.247059

Candidate 2 has the worst multiplier and was removed.

Final order from worst to best: 6, 5, 4, 3, 2, 1. (Notice, it was
unchanged. Not very "chaotic", wouldn't you say?)

On 6/9/12, Kevin Venzke <stepjak at yahoo.fr> wrote:
> Hi Nicholas,
>
> I think that your basic method (page 2 of html version) is the same as
>
> QLTD:
> http://www.mcdougall.org.uk/VM/ISSUE6/P4.HTM
>
> I say this because the multiplier is expressed in terms of ranking slots
> and a candidate is allowed to win with only part of a subsequent slot
> instead of only in increments of entire slots.
>
>
> So your full method is what I would call "QLTD elimination" because you
> repeatedly eliminate the QLTD loser. (Hopefully I haven't misunderstood
>
> the definition.)
>
>
> Elimination+Recalculation methods are bad for monotonicity because the
>
> way information can be used for or against candidates is usually not
>
> predictable. It would need to be quite clear how other candidates will
>
> fare when another candidate is eliminated.
>
> Participation is satisfied by simple point scoring methods. I doubt it is
> compatible with elimination+recalculations. The problem is that you need
> to guarantee each voter that information will only work in certain ways,
> but eliminations tend to have chaotic results.
>
>
> ______________________________
>> De : Nicholas Buckner <nlborlcl at gmail.com>
>>À : Kristofer Munsterhjelm <km_elmet at lavabit.com>
>>Cc : election-methods at lists.electorama.com
>>Envoyé le : Samedi 9 juin 2012 4h04
>>Objet : Re: [EM] Throwing my hat into the ring, possibly to get trampled
>>
>>Thank you for that information. I thought IIA referred to adding of
>>irrelevant alternatives, not removing them. As a consequence I didn't
>>look as strongly at criterions I thought were incompatible, from the
>>Condorcet criterion group.
>
> Basically adding them is a problem if removing them is. If there are only
> two
> candidates A and B and you add a new candidate C, and change the winner from
>
> A to B, then you could also take the new situation, and remove C from it,
> and
> thereby change the winner from B to A.
>
> You wrote originally "I developed an alternative method that takes the
>
> Independence of Irrelevant Alternatives path over the Condorcet path." Do
> you
> know that we don't have *any* serious rank methods that satisfy IIA? For
> example, STV doesn't satisfy it either.
>
> Kevin
>
>