[EM] Re: minmax is not a good public election method

Kevin Venzke stepjak at yahoo.fr
Sun Jun 19 15:25:31 PDT 2005


--- Russ Paielli <6049awj02 at sneakemail.com> a écrit :
> > Hmm, I'm not sure if the CL can win when there is a CW. I should think about
> > that.
> That would be the ultimate embarrassment, eh?

I don't think it is possible. When the CW loses, the MMPO winner has a score
of under 1/2. When the CL wins, in the scenarios I've seen, everybody's score
is over 1/2.

> > Well, it seems to me that when the CW loses in MMPO, it is never very ugly.
> > The MMPO winner would not have any majority-strength loss, for instance.
> That depends on how you define "majority." If you define it in terms of 
> the number of voters who voted a preference in that particular pairwise 
> race, then *every* pairwise loss is a "majority-strength" loss. We've 
> been over this before, of course. You could also define a "majority" in 
> terms of all registered or all eligible voters, but why? What is the 
> fundamental difference between someone abstaining from the entire 
> election or from one particilar pairwise race? And why should one be 
> counted and not the other in your definition of "majority"? (That's 
> intended as a rhetorical question, by the way.)

Why, because you assume it can't be answered?

Someone who abstains from the election evidently doesn't care very much
about the outcome. Someone who abstains from one pairwise contest does so
for reasons we can't know. When these latter voters cause there to be an
absence of a majority-strength win, I don't consider the winner of this
contest to be "cheated" somehow; he could have won, he just needed more votes.
We don't know if he's *really* the majority's preference in the contest.

> >>As most of you realize, we have a dilemma here. You can design an 
> >>election method that counts sincere votes in a reasonable way, or you 
> >>can design one that provides little or no incentive to vote insincerely, 
> >>but you can't do both at once. You want FBC and/or LNH? Then you can't 
> >>satisfy Condorcet.
> > 
> > Well, I've suggested a method or two now which satisfy FBC and come so close
> > to satisfying Condorcet as to make no difference.
> You've proposed some very innovative ideas, and you are obviously a very 
> skilled analyst. However, I don't agree that you can "come so close to 
> satisfying Condorcet as to make no difference."

Really? Note, I was not talking about MMPO when I said that.

> > I prefer Condorcet//Approval with the special tie rule. Maybe I should give
> > it an actual name: "Improved Condorcet Approval." It satisfies FBC, and only
> > fails Condorcet by letting people vote to create pairwise ties. When the CW
> > loses an "ICA" election, he doesn't have a good claim over the ICA winner.
> I don't know how ICA works, but it sounds interesting. Do you mind 
> explaining it or pointing me to its definition? Thanks.

Sure: Assume the use of a ranked ballot, on which all ranked candidates are
considered approved. Find the set S containing every candidate X such that for
any other candidate Y, the number of voters ranking X over Y, plus the number
of voters ranking X and Y tied at the top, is greater than or equal to the number
of voters ranking Y over X. If this means that S is empty, then let S contain all
the candidates. Elect the most-approved candidate in S.

Suppose the winner is A, but the CW is B. If the B supporters argue that B
should have won and not A, the ready response is that the voters ranking A and
B tied at the top intended that A and B have a pairwise tie (and they had the
numbers to make it happen). Also, A must have had higher approval than B.

Kevin Venzke


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