No subject

Thu May 8 11:58:05 PDT 2014

Mainly I wanted to show an example of the count, but also I should say that
in this case Condorcet pairs agrees. The IRV winner might agree, but it is
less well defined with a last place tie in round 2 after Winter is

I know Condoret supporters will question the value of doing well in a
3-candidate election, but I think my approach legitimately gives value to
the higher sets while avoiding the need for elimination of IRV which does
the same less successfully. I accept IRV supporter's claim that there IS
value in a strong plurality count - in measuring "core" support. By looking
at all subsets, no candidate can claim to be spoiled by a competitor.

A candidate that wins in this method must be strong in BOTH the plurality
count AND pairwise counts. I think this is a promising idea, even if it may
be impractical for noncomputer counted elections.

I must admit I think I'd be happier with this approach than Condorcet,
although I don't yet know what properties it might have.

One last note - this last-place count doesn't help choose a winner with the
nasty Condorcet case: (AC=49, BC=48, CB=3)
a1) A=49, B=51 - A last-place
a2) A=49, C=51 - A last-place
a3) B=48, C=52 - B last-place
b1) A=49, B=48, C=3 - C last-place

My last-place count gives a tie between B and C with one loss each. I
somewhat like it that my count can't choose between them since each have
some virtue for different voters, although a true method would need to be
willing to be more decisive. (Perhaps looking at loss-margins from each
subset? I'll leave this open.)

Thanks for listening.

Tom Ruen

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