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.
