[Election-Methods] Elect the Compromise (correction)

[Jobst Heitzig]

Hey folks,

it seems I got my numbers wrong when saying there was a group strategy 
equilibrium in Borda...
Because when the situation is
55 C>A>B
45 C>B>A
the 55 can still change the outcome to A by voting
55 A>B>C instead.
Only this is not an equilibrium either. Seems there is no equilibrium at 
all...

However, other scoring methods will work, e.g. the one proposed by Juho 
and the one Forest announced to me privately (in order to keep the fun): 
Vote against one.

Still, all of them have the cloning problem, so we should try to find a 
clone-proof scoring method that will do. I have the impression that the 
following comes near to it:
Ballots are complete rankings without equal rankings, scores are 
determined like this: Option X gets a point for each tuple (V,W) where V 
is any voter and W is any voter who ranked X above that option which was 
ranked last by V.

For example:
55 A>C>B
45 B>C>A
will give
A a score of (0+55)*55=3025,
B a score of (0+45)*45=2025, and
C a score of 55*55+45*45=5050.
Now the A voters could try
55 A>B>C
but this would change the scores to
A: (0+55)*55=3025,
B: 55*55+(0+45)*45=5050, and
C: 45*45=2025,
which is even worse for the A voters.
If they try
28 A>C>B and
27 A>B>C,
they will produce the scores
A: (27+28)*28+(28+27)*27=3025,
B: 27*27+(0+45)*45=729+2025=2754,
C: 28*28+45*45=784+2025=2809.
So this would produce A, but the B voters can easily react and vote
45 C>B>A
and thus change the result back to C.
However, this is still no equilibrium.

Let's see whether the following is an equilibrium:
55 C>A>B
45 C>B>A
Scores:
A: 55*55=3025
B: 45*45=2025
C: (55+45)*55+(45+55)*45=10000
If now the A voters try
28 A>C>B and
27 A>B>C,
they will change the scores to
A: (27+28)*28+(28+27)*27=3025,
B: 27*27+45*45=729+2025=2754,
C: 28*28+28*45+45*45=784+1260+2025=4069,
so that won't work, and is seems also no other strategy will change the 
winner from C to A.
So it seems this is a group strategy equilibrium. Next, we will have to 
test whether there are other equilibria that elect A or B, and if not, 
whether this one is globally attractive. If not, I guess the scores have 
to be adjusted further...

Yours, Jobst


raphfrk at netscape.net schrieb:
> Forest W Simmons wrote:
> > Below where I said "unlike Borda" I should have said "unlike D2MAC."  
>
> > Neither the Borda solution nor the reverse plurality solution requires 
>
> > anything other than the ordinal preferences.
>
> > 
>
> > So the deterministic solutions that do not depend on some form of vote 
>
> > trading are insensitive to whether or not the voters are inclined to 
>
> > approve C.
>
> > 
>
> > Perhaps we could refine the challenge to ask for methods that elect C 
>
> > with near certainty when the two factions are
>
> > 
>
> > 55 A 100 C 80 B 0
>
> > 45 B 100 C 80 B 0
>
> > 
>
> > but almost surely do not elect C with when the two factions are
>
> > 
>
> > 55 A 100 C 20 B 0
>
> > 45 B 100 C 20 B 0
>
> > 
>
> > assuming throughout optimal strategical voting under near perfect 
>
> > information.
>
> > 
>
>
>
> What about using a clone elimination stage to allow Borda be used.
>
>
>
> For example, select 3 candidates using a PR method.   This is basically
>
> a clone elimination stage.
>
>
>
> The winner is then determined using Borda where all 3 candidates must be
>
> ranked.
>
>
>
> The problem is that A can still be cloned into A1 and A2 as the A faction has a majority and
>
> so can ontain a majority under any reasonable PR method.  However, at least there would be 
>
> a choice between A1 and A2.  The B faction would get to decide which of the 2 A's would 
>
> win.  The end result could easily be that C is A2 (i.e. the 2nd candidate from the A faction 
>
> ... which isn't likely to be infinitely cohesive).  
>
>
>
> Also, the more candidates that are passed through the first stage the better.  If 4 candidates
>
> were passed, then a faction with 40%+1 of the vote can get 2 candidates to the 2nd stage.
>
>
>
> This would yield
>
>
>
> A: 2
>
> B: 2
>
>
>
> If A and B factions both ran clones, then the result is:
>
>
>
> 55:  A1>A2>B2>B1
>
> 45:  B1>B2>A2>A1
>
>
>
> A1 gets 3*55 = 165
>
> A2 gets 2*55 + 45 = 155
>
> B2 gets 55 + 2*45 = 145
>
> B1 gets 3*45 = 135
>
>
>
> This assumes perfect strategy.  The A faction is exactly countering
>
> each B faction vote.  However, it is unlikely that the B faction could
>
> coordinate all to vote the same way without the A faction winning.
>
>
>
> It still doesn't solve the problem, but at least it gives the voters more choice and they 
>
> aren't likely to be party fanatics in general.  Also, if there are lots of candidates
>
> through the first stage, maintaining party uniformity would be alot harder.
>
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