[EM] Net Wins Tiebreakers, Supp. 5
DEMOREP1 at aol.com
DEMOREP1 at aol.com
Wed Jun 2 19:13:32 PDT 1999
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).
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