[EM] An ABE solution

[fsimmons at pcc.edu]

----- Original Message -----
From: Chris Benham 

> Forest,
> 
> "Furthermore something must be wrong with the quoted proof (of 
> the incompatibility of the FBC and the 
> CC) because the winner of the two slot case can be found 
> entirely on the basis of the pairwise matrix."
>  
> The likely explanation for some odd remarks by you and Jameson 
> has just occurred to me. Could it be
> that you and Jameson have mistaken a set of ballots for a 
> "pairwise matrix" ??

The pairwise matrix can be determined from approval ballots in exactly the same way that it is from 
ratings with more slots or from ordinal ballots:

>From each ballot a pairwise matrix is constructed, and then the pairwise matrix for the elcetion is the 
sum of all of these ballot pairwise matrices.

In particular, the entry in row x and column y in the matrix determined by ballot B is a one if ballot B 
approves x and does not approve y.  Otherwise the entry is zero.

>  
> Here is that quoted proof again, with the ballots represented in 
> the more familiar EM notation:
>  
> Hello,
> 
> This is an attempt to demonstrate that Condorcet and FBC are 
> incompatible.I modified Woodall's proof that Condorcet and 
> LNHarm are incompatible.
> (Douglas R. Woodall, "Monotonicity of single-seat preferential 
> election rules",
> Discrete Applied Mathematics 77 (1997), pages 86 and 87.)
> 
> I've suggested before that in order to satisfy FBC, it must be 
> the case
> that increasing the votes for A over B in the pairwise matrix 
> can never 
> increase the probability that the winner comes from {a,b}; that 
> is, it must
> not move the win from some other candidate C to A. This is 
> necessary because
> if sometimes it were possible to move the win from C to A by 
> increasingv[a,b], the voter with the preference order B>A>C 
> would have incentive to
> reverse B and A in his ranking (and equal ranking would be 
> inadequate).

> I won't presently try to argue that this requirement can't be 
> avoided somehow.
> I'm sure it can't be avoided when the method's result is 
> determined solely
> from the pairwise matrix.

This last statement is the one I referred to.  In the case of Approval the winner can be determined soley 
from the pairwise matrix, so Approval satisfies this condition.  It also satisfies the FBC, so the "proof" is 
either wrong or it makes other tacit assumptions that somehow rule out Approval.

On the other hand the three slot version of RCW makes essential use of information beyond the pairwise 
matrix, so even if this proof were correct, it wouldn't apply to n-slot RCW for n>2.

In summary, this "proof"  does not touch RCW.


> 
> Suppose a method satisfies this property, and also Condorcet. 
> Consider this 
> scenario:
> 
> 3: A=B
> 3: A=C
> 3: B=C
> 2: A>C
> 2: B>A
> 2: C>B
> 
> There is an A>C>B>A cycle, and the scenario is "symmetrical," as 
> based on
> the submitted rankings, the candidates can't be differentiated. 
> This means
> that an anonymous and neutral method has to elect each candidate 
> with 33.33%
> probability.
> 
> Now suppose the a=b voters change their vote to a>b (thereby 
> increasing v[a,b]).
> This would turn A into the Condorcet winner, who would have to 
> win with 100% 
> probability due to Condorcet.
> 
> But the probability that the winner comes from {a,b} has 
> increased from 66.67%
> to 100%, so the first property is violated.
> 
> Thus the first property and Condorcet are incompatible, and I 
> contend that FBC
> requires the first property.
> 
> Thoughts?
> 
> Kevin Venzke
>  
> http://lists.electorama.com/pipermail/election-methods-
> electorama.com/2005-June/016410.html
>  
> It is certainly a clear proof of the incompatibilty of  the 
> Condorcet criterion and Kevin's later
> suggested "variation" of  the FBC, "Sincere Favorite":
>  Suppose a subset of the ballots, all identical, rank every 
> candidate in S (where S contains at least two candidates) equal 
> to each other, and above every other candidate. Then, 
> arbitrarily lowering some candidate X from S on these ballots 
> must not increase the probability that the winner comes from S.
> A simpler way to word this would be: You should never be able to 
> help your favorites by lowering one of them.
>  
> http://nodesiege.tripod.com/elections/#critfbc
> 
> I can't see any real difference between this and regular FBC, 
> which probably partly explains
> why it didn't catch on.
>  
> Chris Benham
>  
>  
> 
> ________________________________
> From: "fsimmons at pcc.edu" 
> Sent: Wednesday, 23 November 2011 9:01 AM
> Subject: Re: An ABE solution
> 
> You are right that although the method is defined for any number 
> of slots, I suggested three slots as 
> most practical.
> 
> So my example of two slots was only to disprove the statement 
> the assertion that the method cannot be 
> FBC compliant, since it is obviously compliant in that case.  
> 
> Furthermore something must be wrong with the quoted proof (of 
> the incompatibility of the FBC and the 
> CC) because the winner of the two slot case can be found 
> entirely on the basis of the pairwise matrix.  
> The other escape hatch is to say that two slots are not enough 
> to satisfy anything but the voted ballots 
> version of the Condorcet Criterion.  But this applies equally 
> well to the three slot case.
> 
> Either way the cited "therorem" is not good enough to rule out 
> compliance with the FBC by this new 
> method.
> 
> Indeed, the three slot case does appear to satisfy the FBC as 
> well.  It is an open question.  I did not 
> assert that it does.  But I did say that "IF" it is 
> strategically equivalent to Approval (as Range is, for 
> example) then for "practical purposes" it satisfies the FBC.  
> Perhaps not the letter of the law, but the 
> spirit of the law.  Indeed, in a non-stratetgical environment 
> nobody worries about the FBC, i.e. only 
> strategic voters will betray their favorite. If optimal strategy 
> is approval strategy, and approval strategy 
> requires you to top rate your favorite, then why would you do 
> otherwise?
> Forest