Condorcet(x( ))

[Steve Eppley]

Demorep1 wrote:
>If any group of N voters votes A=B>C, then such votes are obviously
>A N
>B N
>C 0
>A N, B N
>A N, C 0
>B N, C 0
> or
>A N = B N
>A N > C 0
>B N > C 0
>Thus, x = 1.

This isn't obvious to me.  I think you are making some unnecessary
and unstated assumptions about how you think the A=B pairing should
be tallied.  

To me it's more plausible to use x=0 in the A vs B pairing, since
Condorcet penalizes candidates based on "votes against" (votes for
the opponent).  It doesn't reward based on "votes for".  No way would 
the voters who rank A=B as favorites want the method to count 
anything against either one.

>However, the total votes recorded in the A and B combination will
>be more than the number of ballots (which may seem strange to the
>average voter). N votes for A plus N votes for B = 2 times N votes
>recorded. The question is whether or not tie rankings will be
>allowed. If tie votes are not allowed, then there obviously will be
>less voter confusion and the vote totals in any combination will be
>equal to the number of ballots. 

Keeping the totals so they are at most the number of ballots cast can 
be done in other ways than not allowing equal rankings.  As long as 
x <= .5 the totals won't exceed the ballots.

There's no way you can disallow ties unless you disallow truncation.
If two or more candidates are left unranked, they must be treated as 
tied.

--Steve