[EM] pairwise, fairness, and information content
Craig Carey
research at ijs.co.nz
Thu Aug 15 16:46:01 PDT 2002
Solving a 2 candidate election without use of pairwise comparing.
(b0/(a0+b0) < q) implies (B wins)
> q
...
Just to use P1. The candidate under consideration is A:
The same result follows from P2 with no algebra.
 1st 2nd
A a0 a0  u*(1v)
AB ab ab  u*v
B b0 b0 + t + u*(1w)
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] ]
That constraint, "0<=t<=b" is wrong. Instead it ought be
"0<t", or "0<=t<ba". The first of the 2 seems OK.
See, no use of the number of winners. Yet at this list the persons that
write tend to restrict the number of winners to be 1, when it seems far
from obvious how the mathematics of fairness for voters and candidates,
would lead to that. E.g. Richard writes about pairwise comparing but
not how to untangle graphs showing 20 or more winners.
Craig Carey

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