[EM] This might be the method we've been looking for:

[fsimmons at pcc.edu]

Thanks for checking the details.  

In traditional game theory the rational stratetgies are based on the assumption of perfect knowledge, so 
the A faction would know if the B faction was lying about its real preferences.  Even knowing that the 
other faction knew that they were lying they could still threaten to defect, and even carry out their threat.  
There is no absolute way out of that. 

----- Original Message -----
From: Andy Jennings 
Date: Monday, December 12, 2011 12:40 pm
Subject: Re: [EM] This might be the method we've been looking for:
To: Jameson Quinn 
Cc: fsimmons at pcc.edu, election-methods at lists.electorama.com

> You're right. I've drawn out the game theory matrix and the 
> honest outcome:
> 49 C
> 27 A>B
> 24 B>A
> is indeed the stable one, with A winning.
> 
> So the only way for B to win is for his supporters to say they are
> indifferent between A and C and threaten to bullet vote "B". 
> Then the A
> supporters fall for it and vote "A=B" to prevent C from winning. 
> B wins.
> 
> I wonder if this is sequence of events is likely at all.
> 
> ~ Andy
> 
> 
> 
> On Fri, Dec 9, 2011 at 2:31 PM, Jameson Quinn 
> wrote:
> > No, the B group has nothing to gain by defecting; all they can 
> do is bring
> > about a C win. Honestly, A group doesn't have a lot to gain 
> from defecting,
> > either; either they win anyway, or they misread the election 
> and they're
> > actually the B's.
> >
> > Jameson
> >
> > 2011/12/9 Andy Jennings 
> >
> >> Here’s a method that seems to have the important properties 
> that we
> >>> have been worrying about lately:
> >>>
> >>> (1) For each ballot beta, construct two matrices M1 and M2:
> >>> In row X and column Y of matrix M1, enter a one if ballot 
> beta rates X
> >>> above Y or if beta gives a top
> >>> rating to X. Otherwise enter a zero.
> >>> IN row X and column y of matrix M2, enter a 1 if y is rated 
> strictly>>> above x on beta. Otherwise enter a
> >>> zero.
> >>>
> >>> (2) Sum the matrices M1 and M2 over all ballots beta.
> >>>
> >>> (3) Let M be the difference of these respective sums
> >>> .
> >>> (4) Elect the candidate who has the (algebraically) 
> greatest minimum
> >>> row value in matrix M.
> >>>
> >>> Consider the scenario
> >>> 49 C
> >>> 27 A>B
> >>> 24 B>A
> >>> Since there are no equal top ratings, the method elects the same
> >>> candidate A as minmax margins
> >>> would.
> >>>
> >>> In the case
> >>> 49 C
> >>> 27 A>B
> >>> 24 B
> >>> There are no equal top ratings, so the method gives the same 
> result as
> >>> minmax margins, namely C wins
> >>> (by the tie breaking rule based on second lowest row value 
> between B and
> >>> C).
> >>>
> >>> Now for
> >>> 49 C
> >>> 27 A=B
> >>> 24 B
> >>> In this case B wins, so the A supporters have a way of 
> stopping C from
> >>> being elected when they know
> >>> that the B voters really are indifferent between A and C.
> >>>
> >>> The equal top rule for matrix M1 essentially transforms 
> minmax into a
> >>> method satisfying the FBC.
> >>>
> >>> Thoughts?
> >>>
> >>
> >>
> >> To me, it doesn't seem like this fully solves our Approval 
> Bad Example.
> >> There still seems to be a chicken dilemma. Couldn't you 
> also say that the
> >> B voters should equal-top-rank A to stop C from being elected:
> >> 49 C
> >> 27 A
> >> 24 B=A
> >> Then A wins, right?
> >>
> >> But now the A and B groups have a chicken dilemma. They should
> >> equal-top-rank each other to prevent C from winning, but if 
> one group
> >> defects and doesn't equal-top-rank the other, then they get 
> the outright
> >> win.
> >>
> >> Am I wrong?
> >>
> >> ~ Andy
> >>
> >>
> >>
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> >> info
> >>
> >>
> >
>