Dear Jobst, --- Jobst Heitzig <[EMAIL PROTECTED]> a écrit : > you wrote: > > Suppose we're using a method that satisfies Clone-Winner: > > > > 51 A 49 B > > > > A wins. Now replace A with two clones, so > > > > 25 A1>A2 26 A2>A1 49 B > > > > A1 or A2 will win, but only assuming this is how A voters really vote > > after the cloning operation. In real life I suspect B has a very > > good chance of winning. > > I see, very good point! I haven't thought about that before...
This problem worries me a lot, especially if the faction that abandons the other has a better chance of winning: 25 A1 26 A2>A1 49 B > >> Yes, Woodall's definition for the non-deterministic case is > >> different. I was only saying that for deterministic situations both > >> agree! > > > > I don't think this can be true, since a criterion for > > non-deterministic cases can still be applied in deterministic > > circumstances. Revising Woodall's criterion to be deterministic would > > just say that no candidate above the new preference can turn into a > > loser. > > Well, what does Woodalls LNH demand when the method always elects some > candidate deterministically, using randomness only for rare ties? It > demands that when some later preference is added, the winner cannot > change to a less-preferred candidate. > And what does LNPMDO demand when the method always elects some > candidate deterministically, using randomness only for rare ties? It > demands that when some later preference is added, the winner cannot > change to a less-preferred candidate. > That's exaclty the same, they only differ for non-deterministic methods. I hate to press the point, but Woodall's LNHarm can be read exactly the same way for a deterministic method as for a random method. When the vote A>B is changed to A>B>C and this changes the winner from B to A, it's still a LNHarm failure, even if the voter is happy with it. (It's also a Later-no-help failure, since A is assisted. But since the voter probably likes Later-no-help failures, I don't think they are a huge issue.) I don't mean to say that your suggested LNHarm interpretation isn't reasonable. > > I still have some difficulty understanding the sense in which "CW > > else Random Candidate" satisfies a weakened LNHarm. I'll read it > > again. > > I'll try to explain again: (i) When A is the CW, then adding a later > preference to some ballot that already contains A leaves A as the CW, so > nothing changes. (ii) When there is no CW then all candidates have a > positive probability of winning, including my last choice. So when I add > some further preferences to my ballot, then either nothing changes since > there is still no CW, or I produce a CW which will then get elected with > certainty. Since that CW cannot be my last choice, I will have the > positive effect that before the change my last choice was a possible > winner, but after the change it is not. That's what LNPMDO demands. Ok, I understand. That is interesting. > > No, the CDTT can be defined two ways, unless I'm mistaken (and if I > > am mistaken, I hope some list member will correct me!): > > > > The set of all candidates which have majority-strength beatpaths to > > any candidate possessing a majority-strength beatpath to them. (In > > other words, candidate A is in the set unless there is some candidate > > B with a majority- strength beatpath to A, while A doesn't have such > > a beatpath back.) > > Ah, OK, I understand the difference now. The Smith set can be defined > similarly, but using ordinary beatpaths instead of "majority strength > beatpaths", by which you mean a beatpath in which every beat is > supported by more than half of all voters, right? Then it will make a > difference when equal rankings occurr, OK. Right, equal rankings or truncation. It has the property that if A has a majority-strength win over B, adding preferences to ballots for B can't reverse it. That's a vulnerability for Condorcet/Smith, since all pairwise wins are considered. > > The CDTT could be larger or smaller than the Smith set. One reason I > > suggest Random Ballot is that, except for Woodall's DSC method, I > > don't know of any other monotonic, clone-independent, > > LNHarm-satisfying method. > > Do you consider LNH more important than Condorcet-efficiency? Well, in this context, I think it doesn't make sense to pair the CDTT with a method that fails LNHarm, since the point of the CDTT (as I see it) is a good degree of LNHarm. I know of maybe 7 methods that satisfy LNHarm strictly, which is not so many. If I had to choose blindly between a Condorcet method and a LNHarm-satisfying method, I would choose the Condorcet method, since LNHarm alone doesn't seem to guarantee much (consider FPP). I'd insist on the property that if a majority vote X>Y and don't vote Y over anyone (Minimal Defense), then Y mustn't win. The CDTT inherently satisfies this while not strictly satisfying LNHarm. > > I have to admit I haven't put any thought into uncovered candidates > > or their significance. > > I think the significance is only that they possess beatpaths of length > one or two to all other candidates and so majority complaints can be > rebutted most easily. Oh, so that's the goal? Is it mathematically guaranteed that some candidate will have a beatpath of length <=2 to every other candidate? > > I have a dumb question, though. You have a ranking (generated > > randomly, or considering approval, etc.) and an "empty chain." I > > assume the first candidate in the ranking goes into first spot in the > > chain. But what if this is the CW? Then you won't be able to add any > > more candidates. > > There's no such thing as a "dumb" question I guess. What you say is > completely correct. When the CW happens to be first in the sort order, > no further candidate would be added to the chain and the CW would thus > win like s/he should. Ah, I see. That is interesting, if you managed to get a monotone method selecting from the uncovered candidates. > Don't think of the sort order as a ranking, it is only the order in > which I look at the candidates! When there is a CW, no matter where in > the order s/he comes, s/he will win. Only when there is no CW, there are > three or more candidates (=the Banks set) which could be the winner, > depending on the sort order. The significance of the sort order is then > that the later candidates have a better chance of winning since they > need only beat the candidates in the chain, while the first candidate in > the sort order has to beat all other candidates to win. In other words, > TACC elects the least approved candidate only when s/he is the CW, which > is a kind of "intuitive loser" criterion, isn't it? I guess so. No one has proposed an "intuitive loser" criterion, have they? Kevin Venzke Découvrez le nouveau Yahoo! Mail : 250 Mo d'espace de stockage pour vos mails ! Créez votre Yahoo! Mail sur http://fr.mail.yahoo.com/ ---- Election-methods mailing list - see http://electorama.com/em for list info