Discussion:
[EM] Resolvable weighted positional systems all fail independence of clones
Kristofer Munsterhjelm
2017-12-03 20:50:13 UTC
Permalink
A weighted positional system gives x_1 points per first preference, x_2
points per second, etc... down to x_n for last. Usually x_1 >= x_2 >=
... >= x_n.

(Almost) without loss of generality, say that a weighted positional
system gives one point for first preference and zero for second in the
two candidate case. The only method this doesn't cover is the one that
gives equal points to every preference (and thus every election is a
tie), and methods that give higher scores to candidates ranked lower (in
which case, proceed as usual but imagine all the ballots are flipped
around in the examples below).

Then, again AWLOG, let it give one point for first preference, b points
for second, and zero for third in the three-candidate case, 0 <= b <= 1.

Then consider an election where A has a slight majority in the two
candidate case, and x > 0 is some number:

x + 1: A>B
x: B>A

A must win here since it gets x+1 points vs B's x points.

Now clone B into B1>B2 and we get

x + 1: A > B1 > B2
x: B1 > B2 > A

B1's score is the greater of {B1, B2} since 0 <= b <= 1; and B1's score
is (x+1)*b + x = bx + b + x, while A's is x + 1. If we can get B1 to
have greater score than A, then the method will have failed independence
of clones (by teaming).

This happens if

bx + b + x > x + 1,
i.e.
x > 1/b - 1

As long as b > 0, we can always find an x for which this is true.

So suppose b = 0, i.e. that the method is Plurality. Then we can clone A
into A1 and A2 with the following setup:

x/2 + 1: A1>A2>B
x/2 : A2>A1>B
x : B>A1>A2

and since x > x/2+1 > x/2, B wins, so the method fails independence of
clones (by vote splitting).

-

In particular, we can find the range of values for b for which the
method is susceptible to teaming or vote splitting.

Consider the following cloning of A:

x/2+1: A1>A2>B
x/2: A2>A1>B
x/2: B>A1>A2
x/2: B>A2>A1

B's score is x. A1's score is the greater of the two As because of the
single extra vote for A1>A2>B, and A1's score is

x/2 + 1 + bx,
so we have vote splitting when x > x/2 + 1 + bx. This happens for b <
1/2 (then letting x > 2 / (1 - 2b)), so vote splitting happens for b =
[0, 1/2). Teaming happens for b = (0,1].

-

All of this makes it unlikely that I can make a cloneproof extension of
the fpA-fpC method simply by replacing "first preference" (in fpA and
fpC) with some weighted positional system.
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