[EM] MJ: Worse Chicken Dilemma than Approval or Score, elaborate bylaws, computation-intensive count.

[Jameson Quinn]

Two-level MJ is approval, because of the tiebreaker.

Example: Say A gets 52% approval and B gets 57%. Both will have a median of
"approved". After removing 4% "approved" votes from each, A's median will
drop to "unapproved", and B will win.

So if probabilistic SFR works in approval, it works in two-level MJ. And it
also works in pure-100%-strategic MJ. And also for a divided majority, it
works to use probabilistic SFR using grades of min and min+1.

Until you understand that, this discussion is going nowhere.

Also, the first time in this thread that I said CMJ, I linked to the
electowiki page which describes it. As I've told you to do many times.
Here's the link again: http://wiki.electorama.com/wiki/index.php?title=CMJ

If you continue to insist on the same points, without actually listening, I
won't respond.

2012/9/7 Michael Ossipoff <email9648742 at gmail.com>

> I'd said:
>
> >> ...and it [SFR] isn't available in MJ, for the reasons that I described
> in
> >> my previous reply to you.
> >
> Jameson replied:
>
> > Yes it is. Because with approval-style votes, MJ gives approval results.
>
> No. Not with a different count-rule.
>
> I'm just guessing, but you seem to want to say that, though
> Score-style SFR won't work in MJ, Approval-style probabilistic SFR
> will work. Ok, let's look at what would happen:
>
> Suppose that you want to do probabilistic SFR in MJ. You want to
> probabilistically give N points to candidate X. So, with a probability
> of N/max, you give X max points instead of 0 points, as you would in
> Approval, to probabilistically give N points to X.
>
> What will be the result?:
>
> Depending on what N/max is, and depending on the sizes of the
> factions, on on how other factions vote, X's MJ score might be 0, or
> max, or some inbetween amount that you and your faction have _no_
> influence on.
>
> In other words, probabilistic SFR doesn't work in MJ, just as
> Score-style SFR doesn't work in MJ (as I showed in previous postings)
>
> Jameson, you really need to better say what you mean. You need to
> better specify whatever strategy it is that you want to suggest for
> MJ, in order to achieve SFR.
>
> I've told you why Score-style SFR won't work in MJ, and I've just now
> told you why Approval-style probabilistic SFR won't work in MJ. If
> there's some other strategy that you think can achieve SFR in MJ, then
> you need to actually specify it. And then, don't forget to furnish an
> example to show that your strategy works, and how it works.
>
>
> > if it's possible under approval, it is possible under MJ.
>
> Certainly not. MJ isn't counted as Approval is counted. I've just told
> you what would happen if you attempted Approval-style probabilistic
> SFR in MJ.
>
> > And in scenarios like the one you gave, where the median of the unified
> > minority candidate (C) is known (0 in your case)
>
> Sure, C is known to have a median of 0, provided that A voters and B
> voters add up to a majority, and 0-rate C.
>
> > , it doesn't require votes
> > of max or min; it can be done just as well with votes of min or min+1
>
> ...in order for A voters to help B to beat C. Certainly. The problem
> is that if the B voters don't reciprocate, and give 0 to A, then B
> will win by defection.
>
>
> >> SFR could be done unilaterally, or could be done by agreement--an
> >> agreement that doesn't depend on trust, but only on the other
> >> faction's self-interest.
> >>
> >
> > As in MJ.
>
> No. I've told why Score-style SFR, and Approval-style probabilistic
> SFR, won't work in MJ.
>
> You can't have MJ with Score properties. You have to choose between MJ
> and Score.
>
> >> Several people at EM have discussed and demonstrated why Approval soon
> >> homes in on the voter median, and then stays there.
> >
> >
> > Did you even read that page? Because that's a non-sequitor response to
> that
> > page, as far as I can tell. You're just repeating prior assertions.
>
> You'd said something to the effect that median does well, or the
> median does well. I assumed that you meant that the median candidate
> does well in Approval and Score. But apparently you were temporarily
> changing MJ's name to "median". Ok, that's fine.
>
> So you're saying that a more sophisticated discussion at a website
> shows that MJ does well, whatever that means.
>
> That's nice, but I've shown here that MJ doesn't do SFR at all.
>
> How regrettable that you're unable to quote those highly sophisticated
> arguments here, from that website. So then, you're saying that that
> website's more sophisticated arguments show that MJ can do SFR? Or, if
> that isn't what you mean, then do you want to tell us what you mean?
> If you don't want to, that's ok.
>
> Yes, by all means, if you want to, do quote for us those more
> sophisticated arguments that show that SFR can be done in MJ, or that
> the Chicken Dilemma is less serious with MJ.
>
> But "handwaving" at a website just won't do.
>
> ...a vague statement that some website's sophisticated arguments show
> that MJ does well, whatever that means.
>
> >
> >>
> >> In fairly recent postings, I've told some reasons why the Chicken
> >> Dilemma won't be as much of a problem when looked at over time (as
> >> opposed to in one single particular election) in Approval or Score.
> >> But sometimes one wants to avoid the Chicken Dilemma in one particular
> >> election. That's when SFR is more important. But it's helpful in
> >> general too--and unavailable for MJ.
> >
> >
> > Wrong.
>
> So you've claimed. I've told why Score-style SFR won't work in MJ.
> And, in this post, I've told why Approval-style probabilistic SFR
> won't work in MJ.
>
> >> > In practice, in MJ both factions could rate each other's candidate at
> 1
> >> > (the
> >> > second-from-bottom rating). This would mean that any further defection
> >> > would
> >> > be risky
> >>
> >> No it wouldn't. If the A voters rate B at 1 (out of 100), and the B
> >> voters rate A at 0, then here are the MJ scores:
> >>
> >> A: 0
> >> B: 1
> >> C: 0
> >>
> >> (...for the reasons described in the post before this one, the post
> >> that you're replying to)
> >>
> >> B wins by defection.
> >
> >
> > This defection is dangerous: if both sides do it, C wins.
>
> Exactly. That's a necessary condition to have a Chicken Dilemma.
> Because bilateral defection is so dangerous, the A voters, being more
> co-operative, feel compelled to not defect. And that's why they're had
> by the B voters.
>
> > And it is not a
> > temptation as with score or approval: unlike score or approval, it is
> > impossible for defection short of that required to give C a chance, to
> give
> > A or B an advantage.
>
> You need to re-word that, to better say what you mean (provided that
> you mean something and know what you mean).
>
> Of course defection by B "gives C a chance" if the A and B factions
> both defect. That's why there's a Chicken Dilemma. That's equally true
> in MJ, as you yourself agreed in some abovequoted text.
>
> >> Sure, if the A voters and the B voters both give eachother's candidate
> >> a point, then the winner will be A or B. But that just means that
> >> there isn't a problem if no one defects. The Chicken Dilemma is about
> >> what happens when someone _does_ defect.
> >
> >
> > Yes. When some ONE. Not when some entire faction, as in your example.
>
> It's customary, when speaking about such problems as the Chicken
> Dilemma, to speak of there being two "players". That doesn't mean that
> there are only two voters in the election. It means that the A and B
> factions are each collectively referred to as a "player". That's a
> convenient simplification.
>
> Are you saying that, in a large election, MJ doesn't have a defection
> problem if only one voter defects?
>
> > You don't understand MJ or CMJ.
>
> I have no idea what CMJ is. I've been talking only about MJ, because
> it's a popular proposal.
>
>
> > They both have "tiebreaking" procedures that
> > would naturally give the right result.
>
> Of course. MJ needs that. Do you remember when I said that MJ has
> elaborate bylaws?
>
> >> For one thing, tie-proneness isn't
> >> considered a good property.
> >
> >
> > In CMJ, the "tiebreaker" is an integral part of the process, such that
> the
> > tie is broken before it even exists. There is no sense in which CMJ can
> be
> > called tie-prone.
>
> Again, I have no idea what CMJ is. But of course, it goes without
> saying that when the needed tiebreakers are added to a tie-prone
> method, then it can be called "not tie-prone".
>
> When MJ gives the same median score to two candidates, as in the
> example I discussed, and if you wouldn't flip a coin--you forgot to
> tell us what you'd do instead. But it doesn't matter. It comes under
> the term "elaborate bylaws".
>
> >> For another thing, A's win probability
> >> will be 1/2, even if A has many more voters than B has.
> >
> >
> > Wrong.
>
> Jameson, when you say, "Wrong", you should then tell why it's wrong :-)
>
> Ok, if, in the situation that I described, where the A and B factions
> both rated eachother's candidate at 1, and they both ended up with
> equal MJ scores: If you wouldn't flip a coin to choose between those
> two equal-MJ-score candidates, how would you choose between them? In
> MJ. We're talking about MJ, as it's popularly proposed, not CMJ,
> whatever that is.
>
> >> If you want to talk about co-operative trust (as you were doing),
> >> then, in Score, each faction could agree too trustingly and ethically
> >> give eachother's candidate max minus one. Then, they're helping
> >> eachother nearly maximally against C, and yet whichever of {A,B} has
> >> more voters will be the winner.
> >>
> >> That's another thing that won't work in MJ.
> >
> >
> > Um, yes it will.
>
> Um, if both the A and B factions give the same non-0, non-max, rating
> to eachother's candidate, and if neither A nor B has a majority voting
> it 0 or max, and if the C voters give 0 to A and B, then A and B will
> both have the same MJ score. For reasons that I've already told.
>
> In other words, even if the A faction is larger than the B faction, A
> and B will still have the same MJ score. In other words, the
> co-operative strategy that I described doesn't work in MJ.
>
> Um?
>
> Mike Ossipoff
>
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