[EM] Cardinal systems: A generalization of STAR
Kristofer Munsterhjelm
km_elmet at t-online.de
Tue Oct 19 05:10:32 PDT 2021
Here's a generalization of STAR inspired by my ponderings on utilitarian
methods.
Each voter provides a preference ranking, as well as indifference
lotteries over all three-tuples. These indifference lotteries are of the
type: "If only A, B, and C were running, then I would consider B winning
with certainty to be equal in value to A winning with probability p, C
winning with probability (1-p)". The example lottery here is consistent
with either the ranking A>B>C or C>B>A.
Now, the method class. It's parameterized by an internal cardinal
method. For simplicity's sake, let this be Range here. Then the method
goes like this:
If there are only two candidates, the majority candidate wins. So this
is a type-two method - it doesn't handle pizza elections.
For three or more candidates, there's a function that takes a group of
three candidates and returns the loser. It determines the loser by
conducting a Range election with only those three candidates, and with
each ballot being normalized. The loser of the resulting Range election
is the loser that the function provides.
To be more precise about the normalization, suppose that a voter voted
A>B>C and considers a 70% chance of electing A, 30% chance of C to be
equal to electing B with certainty. Then the ratings are chosen so that:
- 0.7 times A's rating + 0.3 times C's rating = B's rating
- A is rated 1 and C is rated 0 (using the full range of the ballot).
In other words, A is rated 1, B is rated 0.7, and C is rated 0. In
general, the "certain candidate" (the middle one in the ranking) is
rated p, where p is the probability of electing A in the indifference
lottery, since p*1 + (1-p)*0 = p.[1]
Finally, let the potential winner set be the smallest set of candidates
so that every Range election with two from the set as well as someone
outside, makes the one outside lose. Run a conventional Condorcet method
on the members of the potential winning set and the provided rankings to
determine the final winner.
Does this distinguish between awful and good centrists while only using
lottery information? Yes. LCR with a good centrist:
40: L>C>R, p = 0.8
30: R>C>L, p = 0.8
20: C>R>L, p = 0.5
The internal Range election goes like this:
L = 40*1 + 30*0 + 20*0 = 40
R = 40*0 + 30*1 + 20*0.5 = 50
C = 40*0.8 + 30*0.8 + 20*1 = 76
The loser is L, so the PW set is {C, R}, and the C vs R election is
decided in favor of C.
Bad centrist:
40: L>C>R, p = 0.1
30: R>C>L, p = 0.1
20: C>R>L, p = 0.5
L and R's scores are as above, and then
C = 40 * 0.1 + 30 * 0.1 + 20*1 = 27
The loser is C, so the PW set is {L, R}, and the L vs R election is
decided in favor of R.
Now, there may still be an incentive to strategically misreport. In
particular, it doesn't solve the three-candidate Burr dilemma. Forest
suggested that Euclidean normalized cumulative voting would fix the
problem because the optima would not be found at the corners of the
polytope, i.e the optimal strategic vote doesn't rate everybody either
max or min.
But I'm not sure just how to do the normalization. Standard L2
normalization involves subtracting the mean and dividing by the standard
deviation; but AFAIK, quadratic voting proposals constrain all weights
to be positive.
The former gives a constrained setup like:
||v||_2 = 1
pv_A + (1-p)v_C = v_B
v_A >= v_B >= v_C
for an A>B>C voter with probability p on the indifference lottery, and
output rating values given by the vector v; but this still has free
variables.
The latter gives:
||v||_2 = 1
v_B = pv_A
v_A >= v_B >= v_C nonnegative
which gives the solution
v_A = 1/sqrt(p^2+1),
v_B = pv_A
v_C = 0.
-km
[1] Note that a completely indifferent ballot can be ignored, since it
would add the same amount to every candidate's score. Thus we don't
actually need to specify what value such a ballot should have. Dealing
with equality, e.g. A=B>C, is harder, and I haven't done that here, but
in the limit, p->1 for an A=B>C ballot (indifferent between certainty
that A gets elected and certainty that B gets elected) and p->0 for an
A>B=C ballot. I would think.
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