[EM] Smith, FPP fails Minimal Defense and Clone-Winner (was "Burlington Vermont repeals IRV 52% to 48%")
cbenhamau at yahoo.com.au
Tue Mar 9 20:45:05 PST 2010
Robert Bristow-Johnson wrote (5 March 2010):
> i like Ranked Pairs best, too. and if the Smith Set are three
> candidates, it and Shulze pick the same winner.
> >/ Bringing Plurality in would be a distraction, since we have no need
> />/ to go near this method and risk a worse answer.
> /it's a "worse answer" in a weird circumstance where an argument could
> be made that any in the Smith set have
> some reasonable claim or legitimacy to be elected. why not the guy
> with the most votes?
Interpreting "most votes" as 'top-ranked on the greatest number of
ballots', the answer is that the resulting method
fails the Clone-Winner and Minimal Defense criteria.
A>C 49-27, C>B 27-24, B>A 51-49
More than half the voters have ranked B above A and A no higher than
equal bottom, and yet Smith,FPP elects A.
The Ranked Pairs and River and Smith//MinMax and Schulze algorithms,
using Margins as the measure of defeat
strength (referred to collectively as "Margins") and IRV also elect A.
Say we replace A with a set of clones, A1 and A2.
(A2 > C > B > A1 > A2)
Now Smith,FPP elects C violating Clone-Winner. Those other methods I
mentioned meet Clone-Winner and so
elect one of the clones (A1).
The number of people (ballots) that voted for (ranked above
equal-bottom) the clones and not C exceeds the number
that voted for C, and yet C wins violating my suggested "Strong Minimal
Douglas Woodall splits Clone Independence into "Clone-Winner" and
Clone-Winner says that if winning candidate X is replaced by a set of
clones than the winner must come from that
set. Clone-Loser says that the winner shouldn't change if one (or
more) losers are replaced by a set (or sets) of clones.
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