[EM] pairwise, fairness, and information content

Craig Carey research at ijs.co.nz
Thu Aug 15 16:11:46 PDT 2002

At 02\08\14 19:12 -0700 Wednesday, Richard Moore wrote:
 >A couple of weeks ago, Craig Carey made an assertion that (to me, at
 >least) was astonishing. He said that a pairwise comparison of
 >candidates contains no information about the "right winner" (his
 >phrase) in an election. That statement should be examined closely.
 >If Craig's statement were true, then, in a two-candidate election,

                                  ^^^^^ example of no inference

 >pairwise comparing would give the wrong result as often as it gives
 >the right result. That would have far-reaching consequences in the
 >study of election methods.

The definition of monotonicity was wrong. In general (i.e. for some
number of winners and candidates), all 3 rules ought be rejected for
failing a perfect method.

All those other members that saw that can write in.

Above you wrote, "as often as".

I request the formula that is returning the probability value that
text is apparently involved in.

I know you can't answer the question Richard. Did the dog get the
probability function?. Do you want to consult with the friends of the
Election Methods List ?.

I solved the 2 candidate method easily:

A    a0
B    b0

A fair but biased solution is:

(q < a0/(a0+b0)) implies (A wins)
(b0/(a0+b0) < q) implies (B wins)

So ties are perfectly handled with even being explicitly mentioned.

Just to use P1. The candidate under consideration is A:

       1st     2nd
A     a0    a0 - u*(1-v)
AB    ab    ab - u*v
B     b0    b0 + t + u*(1-w)
BA    ba    ba - t + u*w

Define: b=b0+ba, a=a0+ab. A usual case has p=1/2.

(For all a0,ab,b0,ba,t,u,v,w) [
  ((0<=t<=b)(0<=u<=a)(0<=v<=1)(0<=w<=1)) implies
   [A loses the 1st  .implies. A loses the 2nd] ]

I imagined Richard was mindful of that formula or something
equivalent when I wrote that FPTP woule be embedded. It is something
as simple as getting the last preference correctly dealt with. The
members can solve the equation. After all, it is a parameterised 2
winner problem. Richard still has not admitted that there is no
need to use pairwise comparing. It is not in the text above so Richard
either is wrong or will be expecting that the text above is wrong,
unless that dictator idea somehow contradicts.

Craig Carey

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