IIAC and Condorcet
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.
Suppose that M is IIAC compliant and elects X. Let Y be any candidate other
than X.
One by one eliminate every candidate other than X or Y, until X and Y are
the only remaining candidates.
The IIAC compliance of M guarantees that X continues to be the winner after
each removal.
So, in the head-to-head contest between X and Y, candidate X must win.
Since the choice of Y was arbitrary, we have shown that X beats every Y
head to head, i.e. X is the ballot head to head winner.
Suppose, now that X cannot beat Y head to head under method M unless X is
ranked ahead of Y on more ballots than not.
This additional supposition completes the proof that X had to be a ballot
Condorcet Winner, after all.
In summary, we have shown that if X wins under an IIAC compliant method,
then X must be a CW for the ballot set in question.
This is the contrapositive of ...
If an election method M elects some candidate X that is not a (ballot)
Condorcet candidate, then method M is not IIAC compliant.
There are two ways the conclusion can be satisfied... 1. The ballot set had
no CW candidate. 2. The method did not satisfy the Condorcet Criterion.
What are the implications for IIAC compliance simulations?
-
Any CC compliant method will have the same maximal IIA compliance as any
other CC compliant method. -
Any non-Condorcet method will have IIA violations that no Condorcet
method has. -
But Condorcet methods violate the IIA only for ballot sets for which no
method can satisfy the IIA. -
Although all Condorcet methods satisfy the IIA for the same ballot sets
(those for which the Smith set is a singleton), their finish orders may
differ on the question of compliance with Local Independence of Irrelevant
Alternatives.
In sum, it is not necessary to test IIA compliance of Condorcet methods by
simulation ... they pass if and only if there is a ballot CW.
It is not necessary to check IIA compliance on any ballot set for which
there is no CW ... every method fails on every such ballot set ... 100
percent failure. If the simulation doesn't show this, then the simulation
is incomplete and misleading.
There are two things you can learn from a simulation about IIA ...1. how
frequently do various non-condorcet methods satisfy the IIA when possible,
i.e. when there is a ballot CW.... and 2. how frequently does any method,
Condorcet or not yield an LIIA finish order.
Every method whose final step is to sort the finish order pairwise will
have 100 percent compliance with LIIAC, so you don't have to check those.
(But of course you can use them to help debug your simulation.)
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.
Dear All,
While I cannot pretend to your sophistication, I have to say that your approach is perhaps defined as the social choice approach -- No?-- determining a winner(s) or authentic election.
It has something of the American cultural concern of whether a person is "a winner." (I sometimes heard Judge Judy Sheindlin, a national treasure, I admit, using the term.)
In my view, there is no winner(s) to be determined; no authentic elected. There is just best estimates. Elections, as someone said, are a "statistic" not a deduction.
American individualism has got as far as putting ranked choice or preference voting, on the reform agenda. It is still far behind Clarence Hoag and George Hallett, of "Proportional Representation. The key to democracy." It is a case of back to the future.
STV research is a progressive endeavor. Hence, my manuals on Binomial STV.
Regards,
Richard Lung.
On 7 Nov 2021, at 10:05 am, Kristofer Munsterhjelm km_elmet@t-online.de wrote:
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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