Jobst--
As I say in the subject-line, this is just a quick preliminary reply:
You wrote:
I wonder whether I understand the NES stuff right. Let me consider a very simple example: 3 options A,B,C, 3 voters, sincere preferences are complete and strict rankings, base method is plurality.
I reply:
Ok, but remember that no one has proposed NES(Plurality). I propose NES(Approval) and NES(Condorcet(wv)), and NES(NES(Condorcet(wv))), NES(NES(Approval)), etc.
So it would be better to test those methods, starting with the simplest, of course. Starting preferably with NES(Approval).
You continued:
when (b) are the sincere preferences, voting (b) is still not an equilibrium in NES(plurality), but voting (c) is the equilibrium instead!
I reply:
But I never said that NES would violate the Gibbard-Satterthwaite theorem and be completely strategy-free. I found that NES seems to share the properties of Condorcet(wv) to a large extent.
That means that there's an offensive order-reversal strategy, and defensive truncation. It seems to me that defensive equal ranking didn't always work in NES(Approval), as it does in Condorcret(wv).
I've recently suggested that maybe nested combinations of NES could be a place to look for a method that meets SSFC. For that reason of course I thank you for checking NES out.
You continued:
So, when the whole thing should make sense, we would have to consider NES^2(base method) := NES(NES(base method)) and so on... Fortunately, for any combination of finitely many options and voters there is only a finite number of possible voting methods we must take into account. Hence the sequence NES^k must become stable or periodic for large k. If it becomes stable, that is, if NES^(k+1)=NES^k for some k, then it seems that NES^k would be a method in which the equilibrium result always equals the sincere result.
I reply:
Ok, but remember that I don't expect NES or any other nonprobabilistic method to be entirely strategy-free. I don't expect sincere rankings to always be a Nash equiilbrium.
Proof: NES^k(equilibrium votes)=NES^(k+1)(sincere prefs)=NES^k(sincere prefs)? That doesn't mean that NES^k leaves no incentives to vote strategically at all, but that at least the situations attained by voting strategically are unstable and will in the end lead to an equilibrium which gives the same result as the sincere one. But what if the sequence NES^k does not become stable but periodic? Is there a similar argument then?
Another question is: When can we be sure that equilibria exist? With plurality, it seems that the equilibria are exactly those voting situations where a majority votes for the Condorcet winner. In all other situations, some majority would have an incentive to vote for the Condorcet winner instead. Without a Condorcet winner, there would be no equilibria. Are there base methods of which we know that equilibria must exist? Does that follow from some well-known criterion for example?
I reply:
NES's rules deal with situations where there's no Nash equilibrium, or in which there's more than 1 Nash equilibrium.
When I was looking at NES(Approval), either I didn't run across an example in which there were no Nash equiilbria, or I didn't run across an example in which there was more than 1 Nash Equilibrium. I'm not sure it was. Probably the latter. I'm not saying that that's always so.
Mike Ossipoff
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