[EM] extending Myerson's test--more policy positions

[Bart Ingles]

David Catchpole wrote:
> Yeeargh! There are three major types of average- Mean, median and
> mode. [...]

Sorry, I always take "average" to mean "mean".  I didn't expect that to
be ambiguous.

[...]
> > > > I wonder what strategy is optimal under a pairwise system?  Should you
> > > > refuse to rank candidates whom you rate below a certain number?  If your
> > > > favorite is one of two known front-runners with a 50-50 chance of
> > > > defeating the other front-runner, it would seem advantageous to refuse
> > > > to rank anyone whom you would rate below the middle of the scale.
> > >
> > > The optimal strategy when no strategy info is available, for a Condorcet
> > > completion system, is sincere expression of preferences... Anyone dare to
> > > argue? Huh? I would feel that any additional strategy brought about by
> > > extra information would depend on the completion.
> >
> > Suppose you are a member of group A below, with vNM utilities as shown
> > in parentheses:
> >
> > votes
> > ??        A(1.0)  B(0.1)  C(0.0)
> > ??        B(1.0)  A, C(??)
> > ??        C(1.0)  A, B(??)
> >
> > If no information is available, you can proceed as though all groups are
> > equal in size.  Thus in a head-to-head matchup between A and C, you give
> > A a 50% chance of winning, so the utility of that matchup (ignoring B)
> > is 0.5.
> 
> No you don't! No you don't! (Sorry) This is not a system where an A voter
> has no information about other voters. She knows there are two other
> groups of roughly equal size to her own. This is somewhat similar to the

What gives you the idea that the three groups above are roughly equal? 
I don't think I'm doing anything that makes an assumption about the
group sizes.  I probably shouldn't have said "as though all groups are
equal in size", although the statement is technically correct.  I didn't
say they "were equal".

If no information is available about the AC pairing, the only reasonable
assumption is that the probability of A>C is 0.5.  I'm not making any
assumption at all about probabilities for A>B or B>C, since they don't
matter here.

This doesn't sound like a prisoners' dilemma, since the voter's decision
doesn't depend on the other groups' actions.  The C voters (and
remaining A voters) either support B or they don't -- if they do so in
sufficient numbers and/or the B group is large enough, B wins anyway. 
If not, our voter has a chance of stopping B.  In neither case is she
punished by failing to rank B.  The only requirement is that this not
affect the outcome of the AC contest.

Of course I'm talking about truncation here, and not order-reversal. 
Also I don't think we can be talking about the same strategy if you are
not using utilities.


> "two big parties' prisoners' dilemma" I mentioned one time aeons ago while
> I was ranting against Condorcet's tendency to elect "centrists" in a
> situation of absolute sincerity. My argument went, in the most simple
> case, that say there were four types of voters ("parties"), such that
> 
> real preferences (not expressing utilities, but you get the drift):
> 
> a: A>B>C
> b1: B>A>C
> b2: B>C>A
> c: C>B>A
> 
> a>b1+b2
> c>b1+b2
> a+b1>c
> c+b2>a
> 
> If b1 and b2 are uncertain, and a and c had roughly the same
> expectations of b1 and b2 as each other, in some circumstances it would
> be Nash-optimal for a and c to collude to reverse their preferences
> between C and their second choice- and even, to probabilistically choose
> between collusion and defection- A prisoners' dilemma.
> 
> >
> > If B wins, the utility of the outcome is only 0.1.   Why would you ever
> > want to help B win by ranking him sincerely?  The only way I would do so
> > is if I knew that A had less than a 10% chance of defeating C.
> >
> >
> 
> --------------------------------------------------------------------
> "Never ascribe to conspiracy what can be put down to stupidity"- DJ No MC