[EM] (Possibly) new method/request for voting paradoxes. :)
Jobst Heitzig
heitzig-j at web.de
Thu Oct 8 06:12:32 PDT 2009
Dear Michael,
very interesting, I don't think I saw anything like this before.
When trying do evaluate a new method, I always try to check very simple
criteria first, like neutrality and anonymity (obviously fulfilled
here), Pareto efficiency, monotonicity, etc. Concerning the latter two,
I was not able to verify them for your method yet. I think you should
focus on those before checking more complex things like Condorcet
efficiency and so on!
Also, immediately a ratings-based generalization came to mind (using
ratings differences instead of rank differences). Finally, when only
seeking a single winner, you could alternatively build a score for each
option X by adding all entries of the final matrix in rows labeled "XY"
and columns labeled with positive differences.
Yours, Jobst
Michael Rouse schrieb:
> 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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