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

[David Catchpole]

On Fri, 17 Mar 2000, Bart Ingles wrote:

> 
> David Catchpole wrote:
> > 
> > On Fri, 17 Mar 2000, Bart Ingles wrote:
> >
> > > The proven optimal strategy when no strategy info is available is to
> > > vote for candidates for whom the voter gives above-average utilities
> > 
> > As in above-median ;> ? I take it as a basic rule of thumb that the
> > optimal strategy when no strategy info is available, in any game, is
> > unrelated to relative magnitudes of utility. If I'm wrong in this case,
> > could you direct me to the proof in question?
> 
> No, not above-median.  Above average.  Example:  there are four
> candidates in the running, whom you rate 10, 9, 8, and 0.  Your average
> for these is 6.75, so the optimal strategy is to vote for the first
> three.

Yeeargh! There are three major types of average- Mean, median and
mode. The mean is the "average" you refer to- it minimises square
deviations. The median is another frequently used "average"- it minimises
modulated deviations. The mode is some weird-f***ed-up-sh** and I don't
know _what_ its useful purpose is. Anyway, they all can be described as
averages, as well as other similar averages such as
root-mean-square. Anyway... there dooz. Thanks for the clarification.

> 
> Example 2:  There are four candidates rated 10, 2, 1, and 0.  Your
> average this time is 3.25, so your optimal strategy is to bullet-vote.
> 
> This assumes nothing is known about the other voters' preferences, so
> that the probabilities of each candidate being in a tie for 1st place
> are equal.
> 
> 
> The proofs can be found in the Merrill book I mentioned, "Making
> Multicandidate Elections More Democratic", Princeton University Press
> 1988, pp 47-63 and 117-120.
> 
> The following were cited in the bibliography of Merrill's book:
> 
> Weber, R. J. (1977) "Comparison of Voting Systems" New Haven: Cowles
> Foundation discussion paper no. 498A
> 
> Merrill, S. (1979) "Approval Voting: A 'Best Buy' Method for
> Multicandidate Elections?", Mathematics Magazine 52:98-102
> 
> 
> The following discusses approval voting strategy, but I'm not sure if
> there are any proofs:  
> http://www.kellogg.nwu.edu/faculty/weber/papers/approval.htm
> 
> 
> 
> > > 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
"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.
> 
> 

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"Never ascribe to conspiracy what can be put down to stupidity"- DJ No MC