# [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)
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

```