[EM] Condorcet same as Approval with partial info?
Bart Ingles
bartman at netgate.net
Sun Feb 20 21:58:46 PST 2000
We all know that Condorcet and Approval would produce the same results
in a situation where every voter's preferences are know to every other
voter (provided there are no cycles). What about the case where partial
information is known?
The following example shows a statistical dead heat between A and C.
Voter utilities are between 0 and 1. I avoid worrying about cycles by
assuming the examples are symmetrical.
Predicted
Votes Candidate(utility)
-------------------------------
49 A(1.0) B(0.1) C(0)
02 B(1.0) A,C(0)
49 C(1.0) B(0.1) A(0)
If A and C both truncate, they each have a 0.5 probability of winning,
so the expected utility to each of these voters is 0.5. If they rank
sincerely, expected utility is only 0.1. It looks as though voters
should bullet vote just as they would under approval voting.
I had been using this example as my "bad Condorcet" example, but in this
situation Condorcet seems to do the same thing as Approval whether under
full or partial info. A group should truncate whenever the von
Morgen-whatsit utility of the sincere CW is lower than the probability
of that group's favorite defeating its least favorite.
Of course, under Condorcet unsophisticated voters might be tricked into
thinking that sincere voting is their best strategy, rather than
truncating when called for.
Two key assumptions are that each group is homogenous, and that each
side can count on the opposite side truncating. The latter shouldn't be
a problem in this case, since the AB lottery is in both sides'
interests. Given those conditions, I wonder if this is true for all
three-candidate cases?
If the groups are not homogenous, there is the possibility that
sub-groups with low utilities for the sincere CW will use order reversal
to overrule groups who really do like the CW. Example based on the
first:
Predicted
Votes Candidate(utility)
-------------------------------
04 A(1.0) B(0.1) C(0)
45 A(1.0) B(0.9) C(0)
02 B(1.0) A=C(0)
45 C(1.0) B(0.9) A(0)
04 C(1.0) B(0.1) A(0)
Condorcet strategy:
04 AC
45 AB
02 B
45 CB
04 CA
53 AB 47
53 CB 47
49 AC 49 ... either A or C wins
This assumes the two 4-voter groups trust each other. The problem is
something like the prisoner's dilemma -- it may not be feasible with a
single isolated election, but with repeated elections cooperation might
be possible. Even if each group thinks there is only a 50-50 chance
that its counterpart will cooperate, it would be the correct strategy so
long as A and C are in a dead heat and the utility assigned to B is less
than 0.25.
This is because p(A/C) given cooperation between both groups = 0.5 (dead
heat), and probability of cooperation is estimated to be 0.5, so overall
p(A/C) = 0.25. Thus the value of this strategy to the 4 AC voters is
0.25 (assuming failure means that C wins). In the example above,
sincere strategy is only valued at 0.1, so order reversal seems the
correct strategy in this case.
Using Approval voting, the CW would have won with ~92% of the vote.
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