[EM] Net Wins Tiebreaker, Supp. 9

[DEMOREP1 at aol.com]

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.