[EM] Smith, FPP fails Minimal Defense and Clone-Winner (was "Burlington Vermont repeals IRV 52% to 48%")

C.Benham 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.

49: A
24: B
27: C>B

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.

26: A1>A2
23: A2>A1
24: B
27: C>B

(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 
Defense" criterion.

Chris Benham




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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