ECONOMY

donald g. saari 395
desired A ∼ C ≻ B plurality ranking. (So, A and C each receive 3 of the vote, B
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receives 2 .) These normalized plurality tallies require the normalized antiplurality
tallies to satisfy 3 16 ≤ X ≤ 1 2 , 1 8 ≤ Y ≤ 1 2 , 3 16 ≤ Z ≤ 1 2 .TohavetheB ≻ C ≻ A
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antiplurality outcome, let X = 3 16 , Z = 5 16 , Y = 8 . By plotting the (x, y, z) and
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(X, Y, Z) point to create the procedure line, it follows that any supporting profile
allows the six election rankings A ∼ C ≻ B, C ≻ A ≻ B, C ≻ A ∼ B, C ≻ B ≻ A,
C ∼ B ≻ A, B ≻ C ≻ A.
To construct a supporting profile, convert (x, y, z)and(X, Y, Z)tointegervalues
where the integer antiplurality outcomes sum to twice the plurality outcomes. (This
is because with the plurality and antiplurality votes, each voter casts, respectively, one
and two points.) With our example, multiply the plurality outcome by its common
denominator of eight to get (3, 2, 3), and the ( 3 16 , 8
16 , 5 ) outcome by its common
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denominator of 16 to obtain the desired (3, 8, 5). (This approach always works if
the common denominator for the normalized antiplurality tally is twice that of the
normalized plurality tally.)
Now notice from Figure 22.1b that the s -coefficient for each candidate is the difference
between her plurality and antiplurality tallies. So a profile satisfying our
example defines the A, B, andC positional outcomes of 3 + s (3 − 3), 2+s (8 − 2),
and 3 + s (5 − 3). These values are listed next to the appropriate Figure 22.2b vertex.
Compute the a through f integervaluesinFigure22.2b as above; e.g. Ann’s tally
of 3 + 0s =[a + b]+s [ f + c]. Ann has no “second-place votes,” so c = f =0. As
Barb’s plurality vote is 2 = (e + f )and f =0,wehavethate = 2. Similarly, d =3.
It remains to divide Ann’s a + b = 3 first-place votes to satisfy the s -coefficients for
Barb and Connie: the only choice is a = 3. Thus, the eight-voter supporting profile
for the example has three voters preferring A ≻ B ≻ C, two preferring B ≻ C ≻ A,
and three preferring C ≻ B ≻ A.
The missing technical condition is that a candidate’s s -coefficient cannot exceed
the sum of the other two candidates’ plurality votes; e.g. as the s -coefficient comes
from the other candidates’ first-place votes, there must be enough of them.
This approach makes it easy to construct profiles to exhibit all sorts of interesting
behavior. In practical terms, just draw a line in the triangle. Almost any line suffices:
if the endpoints are close to the center, the above conditions will be satisfied. Often
just the positioning of the procedure line suffices to discover conclusions; e.g. in this
manner we now know the following. I leave the proofs to the reader; they just involve
drawinganappropriatelineinanequilateraltriangle:
For any ranking, there is a profile where all normalized positional outcomes agree
and yield the selected ranking. (Select the normalized plurality and antiplurality
points to be the same; the procedure line becomes a point where all normalized
tallies agree.)
Select any ranking for the plurality and for the antiplurality outcomes. There
exists a profile with these outcomes. In fact, select any two different positional
methods and any two ranking: there exists a profile so that the first rule’s
outcome is the first ranking while the second rule’s outcome is the second
ranking.