"Candidates A, B and C all have 8 pairwise wins. D has 7.
Could D still be chosen as the winner by any "reasonable" method? "
Sure. D's 7 pairwise
wins could be by a large enough majority that the "extra" pairwise win that A,
B, and C have over the fringe candidate that beats D by 1 vote makes the "8" vs
"7" irrelevant.
This is kind of why I
don't like any counting method that BEGINS with the pairwise matrix. Some
systems would eliminated D when all of A, B, and C are real dogs that no
majority likes.
Rob brown :Hi, I haven't been around for a good while but some of you may remember me.
I have recently been playing around with some stuff for scoring condorcet elections, and ran into a question that seemed obvious but maybe not:
Is it possible, in any of the Condorcet election methods (beatpath, ranked pairs, etc), for someone who has fewer pairwise wins than another candidate, to win the election regardless?
For instance, say there is no Condorcet winner. Candidates A, B and C all have 8 pairwise wins. D has 7. Could D still be chosen as the winner by any "reasonable" method?
Thanks,
-rob
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