[EM] Net Wins Tiebreakers, Supp. 5

[DEMOREP1 at aol.com]

Mr. Cretney wrote in part in Re: [EM] Head to Head Comparison of Election 
Methods, Tue, Jun 1, 1999 1:50 AM EDT--

   1 0 1
40 X Z Y
25 Y X Z
35 Z Y X

X wins

   3  1  1  3
40 X  Z  Y1 Y2
25 Y1 Y2 X  Z
35 Z  Y1 Y2 X

Z wins
----------------------
D ---

The headings should be 
1  0  -1  and
3  1  -1   -3

Net Wins
     X     Y      Z   Total
X   --  -20   30  = 10
Y  20   --   -50  = -30  Min
Z  -30  50   --   =  20   Max
                             0

Splitting Y into Y1 and Y2 clones

     X      Y1      Y2     Z
X   --    -20    -20    30   = -10
Y1  20    --    100    -50  =  70    Max (tie)
Y2  20   -100   --    -50   =   -130  Min
Z    -30    50      50    --   =    70  Max (tie)
                                              0
   
75 Z > 25 Y1
Note the internal Y1/Y2  100/-100 math.
Note the Y1 + Y2 sum is -60.

However, splitting Y into 3 clones--

        X      Y1     Y2     Y3     Z
X     --     -20   -20  -20    30   = -30
Y1   20     --     100   100  -50  =  170  Max
Y2  20      -100  --    100   -50  =  -30
Y3  20     -100   -100   --   -50  =  -230 Min
Z    -30      50      50      50    --   =  120
                                                          0
Note the obvious Y1/Y2/Y3  100/-100 math.
Note - The Y1 + Y2 + Y3 sum is -90.

Thus, to avoid the clone problem, successively drop the choice with the 
fewest net wins. (e.g. Y3 would lose, then Y2, then Y1).

Check for a Condorcet winner (all positive net wins on a row -- a choice 
defeats each choice head to head) and Condorcet losers (all negative net wins 
on a row -- a choice is defeated by each other choice head to head) after a 
choice is dropped.

Thus, the single winner case may often collapse down to 3 choices in circular 
tie cases (assuming each choice is acceptable to a majority of the voters on 
a YES/NO vote). 

It just happens that in the simple 3 choice circular tie that the choice with 
the fewest number of first choice votes loses (giving the false impression to 
so-called Instant Run-Off (IRO) fans to do the same with 4 or more choices) 
(e.g. choice Y in the starting X,Y,Z example) (noting again for the Nth time 
that the IRO folks seem to be incapable of understanding Condorcet head to 
head math).

Thus- YES/NO, head to head, drop lowest net wins (repeatedly, if necessary).