[EM] Condorcet, Weber and Info
Gervase Lam
gervase.lam at group.force9.co.uk
Sun Mar 7 05:58:01 PST 2004
About a month ago, I made a brief mention of the fact that the Weber
formula can be used to investigate Condorcet methods. I'll use it to look
into five cases.
(1) 0-info with a voter's utilities being A=1 B=0.75 C=0.5 D=0.25 E=0
As this case is 0-info, the formula to calculate the Weber score for a
candidate I against a candidate J is Ui - Uj. Using this results in the
following "Weber matrix":
A B C D E
A - -0.25 -0.5 -0.75 -1
B 0.25 - -0.25 -0.5 -0.75
C 0.5 0.25 - -0.25 -0.5
D 0.75 0.5 0.25 - -0.25
E 1 0.75 0.5 0.25 -
If this were Approval voting then the above matrix would be 'collapsed'
into a single row by adding up each column. A one row Approval ballot
matrix consisting of just 0s and 1s is then found such that when each of
its elements is multiplied with the corresponding element in the
'collapsed' matrix, all of the numbers in the resultant matrix are
positive numbers.
As this is Condorcet, the matrix won't be collapsed. Analogous to finding
the Approval ballot matrix, the next step is to find a Condorcet ballot
matrix.
This could be done by going through every ballot ranking permutation and
creating a pairwise matrix for each permutation. The pairwise matrix that
causes the total of the resultant matrix's elements to be as high as
possible is the ballot matrix.
However, in this case, the ballot matrix obviously turns out to be:
A B C D E
A - 0 0 0 0
B 1 - 0 0 0
C 1 1 - 0 0
D 1 1 1 - 0
E 1 1 1 1 -
This corresponds with the ballot A>B>C>D>E. Therefore according to the
Weber formula, in the 0-info case you should vote sincerely.
(2) 0-info with a voter's utilities being A=1 B=0.75 C=0.74 D=0.25 E=0
This is the same as case (1) except that the utilities of B and C are
extremely close together. Therefore, the "Weber matrix" is slightly
different:
A B C D E
A - -0.25 -0.26 -0.75 -1
B 0.25 - -0.01 -0.5 -0.75
C 0.26 0.01 - -0.49 -0.74
D 0.75 0.5 0.49 - -0.25
E 1 0.75 0.74 0.25 -
The next step is to find the Condorcet ballot matrix. This turns out to
be:
A B C D E
A - 0 0 0 0
B 1 - 0 0 0
C 1 1 - 0 0
D 1 1 1 - 0
E 1 1 1 1 -
This is exactly the same as the ballot matrix in case (1). Therefore
according to the Weber formula, in the 0-info case, if there is just a
miniscule difference between two candidates, you should rank one of those
candidates over the other.
(3a) "Partial info" with a voter's utilities being A=1 B=0.75 C=0.5 D=0.25
F=0
In a poll, 100 voters are asked how they would vote in a forthcoming
Condorcet election in which only those 100 are allowed to and have to vote
in. The results of the poll are as follows:
48: A>B>C>D>E
2: A>C>B>D>E
50: E>D>C>B>A
A B C D E
A - 50 50 50 50
B 50 - 52 50 50
C 50 48 - 50 50
D 50 50 50 - 50
E 50 50 50 50 -
This is practically speaking a 0-info case except for the B v. C pairwise
contest. In that pairwise contest, B has a 48% chance of winning the
contest while C has a 52% chance of winning. In all of the other pairwise
contests, each candidate has a 50% chance of winning each contest.
As this is a not a 0-info case, the full Weber formula of Pi * Pj * (Ui -
Uj) needs to be used. This results in the following "Weber matrix":
A B C D E
A - -0.0625 -0.125 -0.1875 -0.25
B 0.0625 - -0.0624 -0.125 -0.1875
C 0.125 0.0624 - -0.0625 -0.125
D 0.1875 0.125 0.0625 - -0.0625
E 0.25 0.1875 0.125 0.0625 -
The next step is to find the Condorcet ballot matrix. This turns out to
be:
A B C D E
A - 0 0 0 0
B 1 - 0 0 0
C 1 1 - 0 0
D 1 1 1 - 0
E 1 1 1 1 -
This is exactly the same as the ballot matrix in case (1). Therefore,
according to the Weber formula, if one of the candidates is slightly
behind another and you prefer the candidate who is slightly behind, you
should vote rank the slightly behind candidate over the other candidate.
The remaining cases are similar enough to each other that I will dispense
with intermediate explanations.
(3b) "Partial info" with a voter's utilities being A=1 B=0.75 C=0.5 D=0.25
F=0
POLL RESULT:
1: A>B>C>D>E
49: A>C>B>D>E
50: E>D>C>B>A
A B C D E
A - 50 50 50 50
B 50 - 99 50 50
C 50 1 - 50 50
D 50 50 50 - 50
E 50 50 50 50 -
"WEBER MATRIX":
A B C D E
A - -0.0625 -0.125 -0.1875 -0.25
B 0.0625 - -0.0025 -0.125 -0.1875
C 0.125 0.0025 - -0.0625 -0.125
D 0.1875 0.125 0.0625 - -0.0625
E 0.25 0.1875 0.125 0.0625 -
BALLOT MATRIX:
A B C D E
A - 0 0 0 0
B 1 - 0 0 0
C 1 1 - 0 0
D 1 1 1 - 0
E 1 1 1 1 -
=> According to the Weber formula, it is best to vote A>B>C>D>E despite
the fact that B is extremely behind in the B v. C pairwise contest.
(3c) "Partial info" with a voter's utilities being A=1 B=0.75 C=0.5 D=0.25
F=0
POLL RESULT:
50: A>C>B>D>E
50: E>D>C>B>A
A B C D E
A - 50 50 50 50
B 50 - 100 50 50
C 50 0 - 50 50
D 50 50 50 - 50
E 50 50 50 50 -
"WEBER MATRIX":
A B C D E
A - -0.0625 -0.125 -0.1875 -0.25
B 0.0625 - -0.25 -0.125 -0.1875
C 0.125 0 - -0.0625 -0.125
D 0.1875 0.125 0.0625 - -0.0625
E 0.25 0.1875 0.125 0.0625 -
This is an "interesting" case. B has no chance of winning against C.
However, the most optimal ballot matrix for this situation turns out to be:
A B C D E
A - 0 0 0 0
B 1 - 0 0 0
C 1 1 - 0 0
D 1 1 1 - 0
E 1 1 1 1 -
This ballot matrix is the same as all of the previous ballot matrices.
Before writing this post, intuition told me that there may be cases where
order reversal of just a couple of candidates would be 'useful' in some
circumstances. However, a couple of things in combination with each other
have shown otherwise:
(1) If Ui - Uj results in a negative value in the "Weber matrix", then the
process of finding a Condorcet ballot matrix inherently tries to put a 0
in the ballot matrix in order to "blank out" the negative value.
(2) The poll results are too mundane. If they were more complicated, I
think the results would be more interesting.
These mundane examples also show that Utilities don't matter. All that
needs to be known is whether you think one candidate is better than
another. However, I get the feeling that Utilities will matter more the
more complicated the scenario.
Thanks,
Gervase.
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