[EM] (Possibly) new method/request for voting paradoxes. :)
Michael Rouse
mrouse1 at mrouse.com
Wed Oct 7 19:56:37 PDT 2009
As usual with such posts, there is a good chance someone has come up
with the same (or very similar) method, but I thought it had interesting
properties, and was wondering what glaring voting paradoxes it had. In
addition, the number of possible orders is overwhelming if there are a
large number of candidates, and I'm not sure that can be simplified.
Finally, Thunderbird sometimes seems to have weird formatting issues in
email, which may screw up the following into unreadability.
With that in mind, here it is.
Step 1: For each ranked ballot, create a matrix for each pairwise vote,
based on the distance and direction between each candidate. For example,
on the ballot A>B>C, you would get:
-2 -1 0 1 2
AB 0 0 0 1 0
BA 0 1 0 0 0
AC 0 0 0 0 1
CA 1 0 0 0 0
BC 0 0 0 1 0
CB 0 1 0 0 0
Taking the rows in order, this shows that A is one position higher than
B on this ballot, which conversely makes B one position lower than A on
the same ballot. Also, A is two positions above C (C is two positions
below A), and B is one position above C (or C is one position below B).
Such detail may be unnecessary -- simply looking at position 1 and above
is sufficient, if you don't allow ties -- but I wanted to show the symmetry.
Step 2. Add all matrices together. As a simple example, let's consider
the following 12 votes in a circular tie (to make it interesting):
5: A>B>C
3: B>C>A
4: C>A>B
Taking the first line, A>B>C =
-2 -1 0 1 2
AB 0 0 0 1 0
BA 0 1 0 0 0
AC 0 0 0 0 1
CA 1 0 0 0 0
BC 0 0 0 1 0
CB 0 1 0 0 0
Multiplied by 5 gives you:
-2 -1 0 1 2
AB 0 0 0 5 0
BA 0 5 0 0 0
AC 0 0 0 0 5
CA 5 0 0 0 0
BC 0 0 0 5 0
CB 0 5 0 0 0
Taking the second line, 3: B>C>A =
-2 -1 0 1 2
AB 3 0 0 0 0
BA 0 0 0 0 3
AC 0 3 0 0 0
CA 0 0 0 3 0
BC 0 0 0 3 0
CB 0 3 0 0 0
And finally, taking the third line, 4: C>A>B
-2 -1 0 1 2
AB 0 0 0 4 0
BA 0 4 0 0 0
AC 0 4 0 0 0
CA 0 0 0 4 0
BC 4 0 0 0 0
CB 0 0 0 0 4
Adding these together gives you
(5: A>B>C, 3: B>C>A, 4: C>A>B) =
-2 -1 0 1 2
AB 3 0 0 9 0
BA 0 9 0 0 3
AC 0 7 0 0 5
CA 5 0 0 7 0
BC 4 0 0 8 0
CB 0 8 0 0 4
Step 3: This is where a bit of a curve is thrown in. Looking at each
position on the matrix, determine which is less -- the sum of the
numbers above the current position, or the sum of the numbers below the
current position -- and add that number to the value of the current
position. In essence, this is adding the value of a position to the
value of all other positions away from the median. Output that number to
a new matrix in the appropriate spot. Using the numbers above, you end
up with:
-2 -1 0 1 2
AB 3 3 3 9 0
BA 0 9 3 3 3
AC 0 7 5 5 5
CA 5 5 5 7 0
BC 4 4 4 8 0
CB 0 8 4 4 4
Step 4: Looking at each possible preference order, add the values for
the appropriate positions on the matrix, and choose the preference order
with the highest score. Using the above values and excluding ties:
A>B>C = (A>B) [one position] + (B>C) [one position] + (A>>C) [two
positions] = 9+8+5 = 22
B>C>A = 8+7+3 = 18
C>A>B = 7+9+4 = 20
A>C>B = 5+4+0 = 9
C>B>A = 4+3+0 = 7
B>A>C = 3+5+0 = 8
So the ranking of possible orders is
A>B>C = 22
C>A>B = 20
B>C>A =18
A>C>B = 9
B>A>C = 8
C>B>A = 7
A>B>C is the most preferred order. C>B>A is the least preferred order.
****************
As another example (ignoring a few pages of work), here is a set of
ballots used in the Schulze Method entry in Wikipedia:
5:A>C>B>E>D
5:A>D>E>C>B
8:B>E>D>A>C
3:C>A>B>E>D
7:C>A>E>B>D
2:C>B>A>D>E
7:D>C>E>B>A
8:E>B>A>D>C
If I did the math correctly, the winning order is:
E>B>A>D>C = (27+25+30+28)+(23+26+13)+(8+16)+(8) = 204
For comparison, Schulze gives E>A>C>B>D, which this method would score
(22+26+28+19)+(16+17+17)+(0+15)+(0) = 160
(Note, no claim is made as to which one is better, I just wanted to show
the difference.)
**************
I've tried variants using distance multiplied by score (which seems to
encourage strategic raising and lowering of ranks) and absolute ranking
rather than relative ranking (which didn't seem to act as nicely in vote
aggregation), but neither seemed to act as nicely on brief inspection. I
haven't come up with a tiebreaker method yet, either -- like in the
simple example of adding together 1: A>B>C and 1:B>C>A (both orders have
the same score) -- but that can wait.
Any questions, comments, or criticisms (the latter most likely about my
math!) are welcome. Especially welcome would be examples of paradoxes,
and links to the same (or similar) method discussed in the past.
Michael Rouse
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