[EM] Resolvable weighted positional systems all fail independence of clones

Kristofer Munsterhjelm km_elmet at t-online.de
Sun Dec 3 12:50:13 PST 2017

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,
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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