[EM] MinMax variant

[Steve Eppley]

On 7 Mar 2003 at 10:42, Forest Simmons wrote:

> Has anybody ever proposed minimizing the maximum opposition 
> rather than minimizing the maximum defeat?

It's a reasonably good method, although it doesn't satisfy some 
criteria I consider important such as Minimal Defense (which is 
similar to Mike Ossipoff's Strong Defensive Strategy Criterion) and 
Clone Independence. (That's why I think the best method is a 
variation of Ranked Pairs which I call Maximize Affirmed Majorities, 
or MAM.  See the web pages at www.alumni.caltech.edu/~seppley for 
more info and rigorous proofs.  The website is still under 
construction, not yet friendly for people who aren't social 
scientists, and most of the web pages requires a web browser that 
supports HTML 4.0 and Microsoft's "symbol" font to be viewed 
properly.)  

Minimax(pairwise opposition) even satisfies a criterion promoted by 
some advocates of Instant Runoff, which I call "Uncompromising":

   Let w denote the winning alternative given some set 
   of ballots.  If one or more ballots that had only w 
   higher than bottom are  changed so some other
   "compromise" alternative x is raised to second 
   place (still below w but raised over all the other
   alternatives) then w must still win. 

The proof that Minimax(pairwise opposition) satisfies Uncompromising 
is simple:  Raising x increases the pairwise opposition for all 
candidates except w and x, and does not decrease the pairwise 
opposition for any candidate, so w must still have the smallest 
maximum pairwise opposition. 

That criterion can be strengthened somewhat and still be satisfied:  
Changing pairwise indifferences to strict preferences in ballots that 
ranked w top cannot increase w's pairwise opposition or decrease any 
other alternatives' pairwise opposition. 
 
> I know that theoretically this could elect the Condorcet Loser, but it
> seems very unlikely that it would do so.
-snip-

Minimax(pairwise defeat) can also elect a Condorcet Loser. Suppose 
there are 4 alternatives, 3 of them in the top cycle but involved in 
a "vicious" cycle.  If the pairwise defeats of the 4th are slim 
majorities, the 4th wins. 

As for the likelihood, this may depend on how you model voters' 
preferences.  In a spatial model with sincere voting, I think you're 
right.  But what if supporters of the Condorcet Loser vote 
strategically to create a vicious cycle?

Minimax(pairwise opposition) can also defeat a "weak" Condorcet 
Winner, one that wins all its pairings but does not have more than 
half the votes in each of its pairings.  The CW may defeat candidate 
x pairwise by a plurality such as 48% (less than half the votes) and 
that might be x's largest opposition, whereas the CW may have a 
larger opposition, such as 49%, in some pairing.  

But it satisfies a weaker Condorcet-consistency criterion:

   If there is a "strong" Condorcet Winner (an alternative 
   that wins each of its pairings by more than half of the 
   votes) then it must be elected.  

Thus anyone who claims no Condorcet-consistent method satisfies 
Uncompromising is incorrect. 

-- Steve Eppley