[EM] Two/three candidate Borda clone failure rate with strict ballots, and IIA relation
Kristofer Munsterhjelm
km_elmet at t-online.de
Sat Nov 6 13:53:19 PDT 2021
I can think of three ways to count the rate of clone failures: two that
base the count around the "before" scenario (and thus imply adding
candidates), and one that considers pairs of elections without any
reference to which is "before" and which is "after" cloning.
The first is this: election eA demonstrates a clone failure in a method
if there exists a way to introduce a clone to eA, so that the new
election has a different winner from eA, and if the winner of eA was the
candidate that was cloned, the winner in the new election is not the clone.
The second is this: Let the "before" election be eA. Let the number of
elections that can be produced by adding a clone to eA be X. Let the
number of elections where this produces a clone failure be Y. Then eA
contributes a proportion Y/X to the clone failure rate of the method.
(Use probabilities or the measure of the appropriate set difference if
your election sets are uncountable.)
The third is this: Let X be the number of distinct pairs of elections,
one which is among c candidates and another which is among c+1
candidates. Let Y be the number of such pairs (eA, eB) where eA has c
candidates, eB has c+1, eB is identical to eA except one of the
candidates has been cloned, and going from eA to eB demonstrates a clone
failure in the method in question (or conversely, going from eB to eA
demonstrates IIA failure). Then the method's clone independence failure
rate for (c, c+1) candidate elections is the ratio Y/X.
I would think that the first way of counting is the best one, but all
three could be defended. The third one makes the correspondence between
clone failure and a subset of IIA failures most obvious, but I strongly
suspect that any method (even something rife with vote-splitting like
Plurality) would have ratio zero, because the number of eB elections
where some candidate is an exact clone of other candidates is
vanishingly small compared to every pair possible (i.e. the chance that
you'd draw such an election from every possible election of c+1
candidates is essentially zero).
What that means is that, though clone independence removes some subset
of IIA failures, it's an extremely small subset and so should have no
effect whatsoever on IIA failure rate alone. That is, unless the
constraints imposed by clone independence force more failures elsewhere,
but I don't think that happens.
=====
So, since I personally prefer counting method one, that's what I'll use
for Borda. If your simulator uses method two, then the numbers will be
different, but they should still show a nonzero ratio of clone failure.
All clone failure in Borda is due to teaming (strategic entry). So with
full preference orders, two-candidate failures all occur in this
pattern, up to a relabeling of the candidates:
Before:
x: A>B
y: B>A
After:
x: A>{B1, B2}
y: {B1, B2}>A
Since the most effective way of boosting one's Borda score is to have
one's candidate consistently closest to first rank, we can without
losing any clone failures let the "after" scenario be the following:
x: A>B1>B2
y: B1>B2>A
And then there's a clone failure if A wins before but B1 wins after. (B2
can't win because it's ranked behind B1 on every ballot.)
If we go by the convention that last place gives one point (not zero),
then A's score after the cloning is 3x + y. B1's score after the cloning
is 2x + 3y.
So the constraints for clone failure are:
x + y = 1 (We want a fixed number of voters, and since they're all real
numbers, we can arbitrarily set the sum to 1)
x >= y (A must at least tie before the cloning happens)
2x + 3y >= 3x + y (B1 must at least tie after the cloning happens).
The solution space is 1/2 <= x <= 2/3, y = 1-x. This lets us divide
voting space into three groups: one where A wins regardless (cloning has
no effect), one with clone failure, and one where A didn't tie before.
Now there are two snags. First, the actual proportion of elections that
exhibit failure will depend on the distribution used to draw the before
elections - e.g. if the voters are normally distributed, the chance of a
(x=1, y=0) election is not the same as that of a (x=0.5, y=0.5)
election. To solve this problem, I'll just use the simplex model
(Dirichlet) so that I don't have to do any probability adjustments.
Second, the area where A doesn't win may still have a clone failure
where B wins, but then A is cloned and A wins afterwards, i.e. swapping
the labels A and B may provide a clone failure. So properly speaking, we
have the following regions:
x < 1/3 : Clone failure is impossible, B wins no matter what
1/3 <= x <= 1/2: Clone failure is possible by cloning A
1/2 <= x <= 2/3: Clone failure is possible by cloning B
x > 2/3 : Clone failure is impossible, A wins no matter what.
Which means that under the simplex model and counting method one, the
proportion of two-candidate Borda elections susceptible to clone
failure, is 1/3.
To check this, I wrote a quick and dirty program to find clone failures
under impartial culture and the simplex model I used (I've added it to
this mail). An example run gives the following results for ten voters:
Impartial culture: 842/2000 failures, rate: 0.421
Simplex model (Dirichlet): 380/2000 failures, rate: 0.19
and for a hundred thousand voters:
Impartial culture: 1994/2000 failures, rate: 0.997
Simplex model (Dirichlet): 655/2000 failures, rate: 0.3275
The impartial culture failure rate tends to one as the number of voters
approach infinity, while for the simplex model it tends to 1/3, which
agrees with the theoretical value.
-km
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