[EMAIL PROTECTED] wrote: >> ...but why 100 iterations?
Forest Simmons wrote: > It is arbitrary. If it's arbitrary -- but still has a significant impact on the result of the election (in the deterministic case, at least) -- I suspect that many people would have a problem with any particular value used. While I may not be able to reasonably predict the exact results from polls, I can probably figure out enough to realize that the lower the value of J, the better the cances for my non-CW favorite to win. Opponents of the CW (if they are able to identify themselves as such prior to the election) may argue for a lower value of J. Since they may make up a majority of the electorate, this presents a real problem. The refinements you mention later may make this point entirely moot, however. Discounting early rounds or utilizing some sort of low pass filter may mean that no matter WHAT value of J is used, my non-CW favorite doesn't have any consistent advantage. > If there is a CW, the method seems to always converge to it fairly > quickly, but suppose that the set is > 48 A(rank=1) B(rank=3) C(rank=4) > 2 B(rank=1) A(rank=2) C(rank=4) > 3 B(rank=1) C(rank=2) A(rank=4) > 47 C(rank=1) B(rank=50) A(rank=51) . > It will take about fifty iterations for the approval cutoff to work its > way down under rank 50 in the C faction ballots. From that point on B is > the winner. So the marble method gives A and B about equal chances of > winning, while the deterministic method makes the CW the sure winner. > (1) Could anybody have predicted this ahead of time based on a random > sample of the ballots? If not, then there would be little incentive to > falsify preferences. In the deterministic case, I think perhaps somebody COULD have predicted this ahead of time based on a random sample of the ballots. My intuition is that (practical) unpredictability only comes into play in the deterministic case when there is an eventual cycle in the outcome (so that the value of J essentially picks one member of that cycle at random). Since in your example the system eventually settles down to just returning B after every round, and does so well before the J=100 cutoff, I think it might be possible to predict that. With input sets that eventually result in (say) A, B, A, B, A, B, etc. cycles, I suspect the unpredictability results from not knowing WHICH of A or B wins for sufficiently large J. > We start giving all candidates equal weight. If they cannot get even one > win (in an hundred tries) with that crutch, then they don't have enough > support to deserve any further chance. This ties back to what I was saying about Procedural vs. Substantive fairness. Consider an extremely polarized electorate with just two parties: 51 A(rank=1) B(rank=x) 49 B(rank=1) A(rank=y) Candidate A wins every time. While that may seem procedurally fair, given that A has an absolute majority of support, think about what happens in multiple elections of the same type (either different districts voting for a multi-member body, or multiple terms spread over time for a single-member body). The candidates from B's party will NEVER be elected -- despite the fact that they have the support of nearly half the population! This is a very real problem in the South-East US, where ethnic minorities have been historically under-represented in office. In fact, this principle is a big part of the logic behind district jerry-mandering. Each individual election may be "fair" in the sense that it's procedurally fair, but the overall results are hardly fair in actual substance (outcome). > But ultimately, if a candidate not in the equilibrium set has a chance of > winning, then there will be some incentive for distorting preferences (it > seems to me). Random ballot (a.k.a "random dictator") gives every candidate with any measure of first-place support at all a chance of winning, yet there is no incentive whatsoever for distorting preferences. So it's definitely possible (though "reasonable" methods may still fit with your intuitions, I'm not sure). > Do the above comments clarify these questions for you? Yes, thanks. > [Detailed explanation snipped] > If you don't like non-standard analysis, you can translate the whole > thing into epsilons and deltas, at the expense of a lot of clutter, not > to mention loss of intuitiveness. That's okay, I'm (mostly) comfortable with non-standard analysis... although the notion of a procedure with a non-standard number of steps makes my head hurt a little bit (it's a bit too close to a "completed infinity" for my liking... though I don't think it introduces any sort of real complications in this case). One concern I had initially (though I'm not entirely sure it applies) is that there might be more than one probability vector P that has the required properties. What do you think about the possibility of more than one equilibrium set? What if using initial weights of (say) 2 resulted in the system converging on a different candidate or set of candidates? I'm not sure such a thing is possible in the system you describe... but independent local minima/maxima are par for the course in other similar systems (training of Artificial Neural Networks comes to mind). -wclark -- Protest the 2-Party Duopoly: http://vote.3rd.party.xoom.org/ ---- Election-methods mailing list - see http://electorama.com/em for list info
