Clones

[DEMOREP1 at aol.com]

In Re: Condorect sub-cycle rule, Sat, Oct 4, 1997 12:18 PM EDT Mr. Schulze
copied an example from Mr. Saari regarding clones----

> [Saari] Assume the following ratings (opinions):
> 
> Apple-1 Apple-2 Apple-3 Chocolate (Ranking)
> 20% Exc(99) Exc(98) Exc(97) Exc(95) A1>A2>A3>Ch
> 20% Exc(97) Exc(99) Exc(98) Exc(95) A2>A3>A1>Ch
> 15% Exc(98) Exc(97) Exc(99) Exc(95) A3>A1>A2>Ch
> 
> 15% Bad(-30) Bad(-40) Bad(-50) Exc(95) Ch>A1>A2>A3
> 15% Bad(-50) Bad(-30) Bad(-40) Exc(95) Ch>A2>A3>A1
> 15% Bad(-40) Bad(-50) Bad(-30) Exc(95) Ch>A3>A1>A2
> 
> If the ballot contains only (Apple-x vs.Chocolate) then the
> Condorset winner is Apple:
> A > Ch 55%-45%
> Apple wins easily.
> 
> But if the ballot contains (Apple-1, Apple-2, Apple-3, Chocolate)
> then the Condorcet winner is Chocolate:
> A1 > A2 65-35
> A2 > A3 70-30
> A3 > A1 65-35
> A1 > Ch 55-45
> A2 > Ch 55-45
> A3 > Ch 55-45
> (Chocolate is the candidate whose worse pairing defeat is smallest.)
> 
> Because a situation can be created where adding "Twins" to the
> ballot alters the outcome, I conclude that Condorset voting fails the
> "Twins" Litmus Test and therefore is not worthy.

D- I note that Chocolate is the Condorcet winner without doing any tiebreaker
math with the clones.
----
Mr. Schulze wrote--
I want to emphasize, that even Smith//Condorcet[EM] fails to
meet this "Twins" Litmus Test.

Example:

(1)120 voters vote Clinton > Dole > Perot.
   105 voters vote Dole > Perot > Clinton.
   75 voters vote Perot > Clinton > Dole.

   Clinton:Dole = 195:105.
   Clinton:Perot = 120:180.
   Dole:Perot = 225:75.

   Clinton has the lowest number of votes against and
   easily wins the elections.

(2)Now, Clinton is cloned and Clinton1, Clinton2, and Clinton3
   are his twins.

   40 voters vote Clinton1 > Clinton2 > Clinton3 > Dole > Perot.
   40 voters vote Clinton2 > Clinton3 > Clinton1 > Dole > Perot.
   40 voters vote Clinton3 > Clinton1 > Clinton2 > Dole > Perot.
   35 voters vote Dole > Perot > Clinton1 > Clinton2 > Clinton3.
   35 voters vote Dole > Perot > Clinton2 > Clinton3 > Clinton1.
   35 voters vote Dole > Perot > Clinton3 > Clinton1 > Clinton2.
   25 voters vote Perot > Clinton1 > Clinton2 > Clinton3 > Dole.
   25 voters vote Perot > Clinton2 > Clinton3 > Clinton1 > Dole.
   25 voters vote Perot > Clinton3 > Clinton1 > Clinton2 > Dole.
   
   Clinton1:Dole = 195:105.
   Clinton2:Dole = 195:105.
   Clinton3:Dole = 195:105.
   Clinton1:Perot = 120:180.
   Clinton2:Perot = 120:180.
   Clinton3:Perot = 120:180.
   Dole:Perot = 225:75.
   Clinton1:Clinton2 = 200:100.
   Clinton2:Clinton3 = 200:100.
   Clinton1:Clinton3 = 100:200.

   Now, Dole has the lowest number of votes against
   and wins the elections.

   Thus, Clinton's twins change the result of the elections
   without being elected.
------
D- A simple case with 3 choices----

N1>N2>N3, all are nearly equal

N1  A>B>C
N2  B>C>A
N3  C>A>B

N1+N3  A/B  N2
N1+N2  B/C  N3  Maximum for B, Minimum for C (note the C diagonal)
N2+N3  C/A  N1

If A is cloned into 3 clones (such as A1>A2>A3>A1) and the N1, N2 and N3
amounts are divided by 3, then the results are the same in the A clone/B and
C/A clone pairings.

Lesson- there is a very slight chance of such clones in a real public election
but a Condorcet tiebreaker must note such possibility.  Such 3 or more clone
types are in a class of special types such as Condorcet first choice majority
winners or last choice majority losers and Smith set choices.

Expanding to a simple 4 choice case with no clones---
M1>M2>M3>M4, all are nearly equal

N1  A>B>C>D
N2  B>C>D>A
N3  C>D>A>B
N4  D>A>B>C

N1+N4           A/B    N2+N3
N1+N2+N4     B/C    N3
N1+N2+N3     C/D    N4    Maximum for C, Minimum for D (note the D diagonal)
N2+N3+N4     D/A    N1

N1+N4           A/C   N2+N3
N1+N2           B/D   N3+N4

 Should D be dropped ?   I note that dropping one choice leaves 3 pairings
which may or may not be in a circular tie.  Implications for 5 or more choices
in a circular tie ???  More to follow.