[EM] IIAC and Condorcet
Richard Lung
voting at ukscientists.com
Sun Nov 7 02:49:43 PST 2021
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 at 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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