[EM] (Possibly) new method/request for voting paradoxes. :)
Michael Rouse
mrouse1 at mrouse.com
Sun Oct 11 09:00:58 PDT 2009
This method always selects the Condorcet winner if there are an odd
number of voters with no ties or circular ambiguities. *With respect to
a particular candidate,* the other candidates must form a single
uniquely-preferred rank order, with a single peak in ranking score at
the (composite) median voter -- if the median voter places B one place
below A, the median voter cannot also rank C one place below A, unless
both B and C are tied. The converse is also true -- if A is one place
above B, C cannot also be one place above B. If you had a rank order
with the Condorcet winner anywhere other than first position, you could
always raise the total score by moving the Condorcet winner up the
ranking, because the median voter would rank the Condorcet winner above
the other candidates. I *think* this means that the final result will be
the Condorcet order (in the absence of ties), but I haven't tried
checking it yet.
Of course, if there are no voters at the median ranking (for example, if
exactly half of the voters ranked A as first place and half ranked A as
last place, with no voters in the middle -- which is why I mentioned "an
odd number of voters" above), you will have a range of ranks with the
same score. For example, combining the votes A>B>C and B>C>A will
result in a tie between those two rankings, even though a case can be
made that B>A>C would be the preferred ranking, if one must be picked.
So there are probably a whole host of paradoxes out there, in addition
to the ones inherent in Condorcet methods. :)
For ratings differences, do you mean on a range-type ballot? That might
be an interesting path to look at. And do you have an example of the
before and after matrices for your positive difference suggestion?
Michael Rouse
PS Of course, it's entirely possible that I made a mistake in reasoning
somewhere, so I would be interested in any counter-examples where the
Condorcet winner (or Condorcet order) did not match the winner in this
system.
Jobst Heitzig wrote:
> 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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>>
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