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

[fsimmons at pcc.edu]

This new method (Top Tier Pairwise Rule modified MinMax) gives a satisfactory resolution to "Kevin's 
Bad MMPO Example:"

49 A
01 A=C
01 B=C
49 B

It yields a tie between A and B.

Shall we call the method MinMax(TTPR)?

----- Original Message -----
From: 
Date: Friday, December 2, 2011 4:12 pm
Subject: This might be the method we've been looking for:
To: election-methods at lists.electorama.com,

> 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?
>