[EM] Net Wins Tiebreaker, Supp. 9
DEMOREP1 at aol.com
DEMOREP1 at aol.com
Mon Jun 14 13:39:25 PDT 1999
I note the general problem with 3 or more choices in a circular tie in a net
wins table- namely, there are 12 obvious possibilities --
All cells of net wins or losses (in the body of a table), Row Sums and Each
choice row
versus
Max Wins, Min Wins, Max Losses, Min Losses in each category
i.e. 3 x 4 =12 combinations
A simple 2 choice net wins table is like
A B Total
A -- N N
B -N -- -N
Example
51 AB
49 BA
produces
A B Total
A -- 2 2
B -2 -- -2
It would seem that any tiebreaker with 3 or more tied choices must collapse
into a simple 2 choice table.
More information about the Election-Methods
mailing list