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. ---- Election-methods mailing list - see http://electorama.com/em for list info