Condorcet(x( ))

[Steve Eppley]

Demorep1 wrote:
>A way to treat tie rankings in Condorcetish methods is to note that
>(a) at any choice level, a voter has a total of one vote and (b) tie
>rankings cover more than 1 choice level.
>
>Example-- A voter votes A> (B=C=D) > E
>The (B=C=D) involves a tie for the 2nd, 3rd and 4th choice levels.
>Since a voter has a total of one vote at any choice level, there is
>  1st   one vote for A
>  2nd  1/3 vote for each of B, C and D
>  3rd  1/3 vote for each of B, C and D
>  4th  1/3 vote for each of B, C and D
>  5th   one vote for E

This isn't "Condorcetish", imho.  Condorcet's method doesn't count
votes "for", which is what you're doing in your example; it counts
votes "against" (i.e., votes for the pair-opponent).  Nor does
Condorcet's give any meaning to absolute rank positions (one vote 
per "choice level", etc.)

In pure pairwise methods, the ranked ballot A>B=C=D>E is treated as
*just a shorthand*.  There are no "choice levels."  The meaning of
this ballot is: A>B, A>C, A>D, A>E, B=C, B=D, B>E, C=D, C>E, D>E
Nothing more, nothing less.  It's just 10 pairwise preference votes.
(More when you count the unranked candidates.  N(N-1)/2 in all.)

If you suggest methods which take into consideration the absolute
rankings, it's no longer a pure-pairwise method.  That may be okay.
But don't forget to consider the strategy opportunities and dilemmas
the methods may open up.  (In other words, don't assume methods don't
create incentives for voters to misrepresent their preferences.)  

And when you propose methods how about also telling us how they fare
on the standards you *and we* care about?  We haven't had any votes
in EM to rank or rate standards, but it's pretty clear that for some
of us eliminating strategies and dilemmas is important. 

--Steve