[EM] RE : Ranked Preference benefits
mrouse1 at mrouse.com
mrouse1 at mrouse.com
Sun Nov 5 06:27:50 PST 2006
Juho wrote:
> My usual example is this:
> 25:A>B>D>C
> 25:B>C>D>A
> 25:C>A>D>B
> 24:D
>
> There are four parties of about equal size. D is the Condorcet loser
> but on the other hand she is two votes short of being the Condorcet
> winner. The other three candidates are badly cyclic and would each
> need 26 additional votes to become Condorcet winners. In this light D
> is not a bad winner after all. Not electing the Condorcet loser is a
> good property in 99.9% of the elections but in the remaining tricky
> cases things may look different.
I ran the example through the Rob LeGrand's election calculator at
http://cec.wustl.edu/~rhl1/rbvote/calc.html and got the following:
Winner: (A,B,C, requires tiebreaker)
Baldwin*
Black*
Borda*
Copeland*
Nanson*
Raynaud*
Schulze*
Small*
Tideman*
Winnder: D
Dodgson
Simpson
Which seemed like an odd result to me. Any idea why Dodgson and Simpson
gave the Condorcet loser?
On a tangential note, as a test I tried it with IFNOP to see if it would
resolve there, and got the following:
25:A>B>D>C
25:B>C>D>A
25:C>A>D>B
24:D>A=B=C
Left hand side is normal comparison (>), right hand side is reverse (<),
number in parenthesis is the difference.
25 AB 25 (0)
25 AC 25 (0)
50 AD 49 (1)
25 BC 25 (0)
50 BD 49 (1)
50 CD 49 (1)
Since D is the Condorcet loser, it is dropped.
AB 0+0+0+25+0+0
AC 0+0+0+0+0+25
BA 0+0+0+0+0+25
BC 0+0+0+25+0+0
CA 0+0+0+25+0+0
CB 0+0+0+0+0+25
This gives A=B=C -- a tiebreaker would be needed with this method as well.
(It's D's voters' fault for not giving a complete preference, and an
election giving D the win would seem to open such methods for
manipulation.) A runoff election between A, B, and C would probably be the
fairest method.
Michael Rouse
mrouse1 at mrouse.com
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