[EM] IIAC and Condorcet
Kristofer Munsterhjelm
km_elmet at t-online.de
Sun Nov 7 02:05:23 PST 2021
On 11/7/21 7:54 AM, Forest Simmons wrote:
> Nothing new, but under-appreciated....
>
> If an election method M elects some candidate X that is not a Condorcet
> candidate, then the ballot set from that election can be used to show
> that method M is not IIAC compliant.
Your point reminds me of how one can determine that LIIA implies ISDA -
by the same reasoning.
You're right; I was just now thinking the same thing about
three-candidate elections. If we use method one of counting (IIA failure
exists beginning in election eA if there exists a candidate that, when
removed, changes who the winner is), then clearly this is the case.
If there's a three-cycle, then we can remove one of the candidates so
that the original winner is beaten by someone else; this is an IIA failure.
If we extend the notion of IIA failure so that we're permitted to remove
any subset of irrelevant candidates, not just one, then by induction the
same thing holds for any number of candidates whenever the method elects
someone who is not a CW.
That happens for every Condoret method when there is a cycle, and
happens for every non-Condorcet method for some fraction of elections
where there's a CW.
So with this method of counting, and with a subset allowed, every
Condorcet method has the same rate of IIA failure, and no non-Condorcet
(deterministic, majoritarian) method has any lower rate of IIA failure.
Importantly, that's *irrespective of its clone independence
performance*, which I suppose fits with my hunch.[1]
However, just to *completely* shore up the argument, someone might
object that being able to remove an arbitrary subset is too powerful.
Would it be possible to argue that a Condorcet cycle of any size can be
used to induce IIA failure by removing only one candidate?
The reason arbitrary subsets is more powerful than single candidates can
be seen by considering Borda cloning again. Suppose the before is:
x: A>B
y: B>A
and we say that it's a clone failure if we can add some number of clones
and the winner changes. Then after cloning we have:
x: A>B1>B2>...>Bn
y: B1>B2>...>Bn>A
Suppose for simplicity that last place gets zero points. Then A gets
xn+y points and B1 gets x(n-1) + yn >= (x+y)(n-1). Thus by making n
large enough we can get B to win as long as a nonzero fraction of voters
originally voted B>A. This would make Borda's clone failure unity under
pretty much *any* distribution, not just impartial culture.
-km
[1] Since IIA failures are clone failures almost nowhere, the direct
impact should be negligible.
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