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

[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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