[EM] Who can't solve 2 candidate elections

[Markus Schulze]

Dear Craig,

I wrote that in the 2-candidate 1-winner case FPP satisfies
e.g. anonymity, neutrality, non-dictatorship, Pareto,
strategyproofness, monotonicity, participation, consistency,
and resolvability. I asked if, in your opinion, FPP doesn't
"solve" the 2-candidate 1-winner case then what does it mean
to "solve" the 2-candidate 1-winner case?

You replied that you don't consider anonymity, neutrality,
non-dictatorship ("basically undefined"), Pareto ("completely
worthless", "obviously rubbish"), strategyproofness
("unpromising") or participation ("undesirable") important.
However, this is not an answer to my question why FPP doesn't
"solve" the 2-candidate 1-winner case in your opinion. To
answer my question why FPP does _not_ "solve" the 2-candidate
1-winner case, you would have to mention a desirable criterion
that is _not_ satisfied by FPP in the 2-candidate 1-winner case.

Markus Schulze

> > [EM] Who can't solve 2 candidate elections
> > Markus Schulze markus.schulze at alumni.tu-berlin.de
> > Fri Jan 7 07:10:56 PST 2005
> >
> >
> > Dear Craig,
> >
> > you wrote (7 Jan 2005):
> > > Well, I didn't underestimate your intelligence when I
> > > expected that you would be perfectly unable to solve
> > > the easy problem of deriving a solution to the 2 candidate
> > > 1 winner election problem.
> >
> > Well, this depends on what you mean with "solving"
> > 2-candidate 1-winner elections.
> >
> > In the 2-candidate case, FPP satisfies all important
> > criteria (e.g. anonymity, neutrality, non-dictatorship,
> > Pareto, strategyproofness, monotonicity, participation,
> > consistency, resolvability). If, in your opinion, FPP
> > doesn't "solve" the 2-candidate 1-winner case, then what
> > does it mean to "solve" the 2-candidate 1-winner case?
> >
> > Markus Schulze
>
>
> The First Past the Post method is NOT the solution of the
> 2 candidate 1 winner election problem.
>
> Suppose the papers are these:
>
>    a0*(A)
>  + ab*(AB)
>  + b0*(B)
>  + ba*(BA)
>  + z0*()
>
> The First Past the Post solution is (of course), this:
>
>    (0=ab=ba).(b0<a0) implies (A wins)
>    (0=ab=ba).(a0<b0) implies (B wins)
>
> All numbers are real (and any can be negative).
>
> The solution I am expecting is, of course, this:
>
>    (b0+ba<a0+ab) implies (A wins)
>    (a0+ab<b0+ba) implies (B wins)
>
> (The "<" might be replaced with "<=".)
>
> You ask "what does it mean .. ?".
> First Past the Post has a solution trapped inside of a thin
> subspace.
>
> You might want to add (0<=a0) or whatever.
>
> There is more than one way to solve the problem. You should
> use important and correct axioms and not use crap like Pareto
> and maybe Participation.
>
> You might want to disclose the "strictly prefer" weighting numbers.
> Every answer you want to say guarantees that monotonicity is failed.
>
> So when you list monotonicity as an axiom, you are hoaxing or
> bluffing.
>
> I know you know about FPP, but you can't use FPP unless you have
> an axiom that allows you to use FPP.
>
> I have an "embedding" axiom, so I can embed FPP.
> Then shadows are cast using 'strict fairness' rules. That fills up
> all of the space. You didn't have a fairness axiom and you didn't
> have an embedding axiom.
>
>
> The Pareto rule is completely worthless. The definition of
> "strong Pareton" on Eppley's website is obviously rubbish (and
> being a believer in pairwise comparing, he is slower at
> spotting problems)
>
> http://alumnus.caltech.edu/~seppley/
>
>    | Strong Pareto:  If at least one voter ranks alternative y over
>    | alternative x  and no voters rank x over y, then x must not be
>    | elected. 
>
> Here are some problems:
> (1) there are no voters. Why is Eppley talking about voters and not
>    counts of ballot papers ?.
> (2) Suppose there is 0.99 or 1.01 (AB) papers, etc.. Why is Eppley
>    comparing against the number 1 ?.
> (3) Suppose the number of winners equals the number of candidates.
>    He simply boldly requires that the number of winners be 1 when
>    it is before-hand required to be 2. Have you got any axiom
>    saying that the number of winners is correct ?. Mr Eppley
>    seems to requires that the number of winners be wrong, and he
>    didn't motivate that mistake.
>
> Also, I assume that "non-dictatorship" truly is basically undefined.
> I am sure that both of us can't define it.
>
> Since you are designing a 2 candidate method, there is no need
> for the anonymity and neutrality rules:
> hopefully none of your axioms will create a failure so there is no
> need to have those rules to remove a problem. 
>
> You didn't define strategyproofness. That sounds really unpromising.
>
>
> You should not copy from Mr Kenneth May.
>
>
> The Participation rule seems to fail my IFPP so Participation
> seems to be undesirable.
>
> why not tell me about the 5 (or more) weighting numbers that go
> into the "strictly prefer" sum. Mr Steve Eppley also keeps the
> same 5 numbers secret. I suppose Mr Eppley kept the 5 numbers
> secret from you ("A over B" = 1*(A) + 2/3*(AB) ...). And you
> jumped to a conclusion and believed you got told the numbers,
> when in fact he was keeping the 5 numbers secret.
>
> You can select the axioms.
>
> Did you see my logic expression of Avy ?. Just like all of
> your Condorcet variants, it had a large number of faces that
> are failed by monotonicity. Every selection of the 5 secret
> "strictly prefer[s]" weighting numbers, will cause the method to
> be failed by monotonicity.
>
> I think Barney and Eppley and D G Saari and who else ??, all try
> to keep the 5 weighting numbers, secret. Mr Lanphier can speak
> truthfully in the first statement.However he does not abandon
> error.
>
> If I asked Mr Gilmour then he might say that Truncation
> Resistance would be used. I myself wuuld use Woodall's Symmetric
> Completion of 1994.
>
> Aren't you supposed to use some pairwise comparing belief or
> something ?. So that it extrapolates to the 3 candidate case?.
>
> So you can't solve 2 candidate elections.