[EM] (Possibly) new method/request for voting paradoxes. :)
Dave Ketchum
davek at clarityconnect.com
Sun Oct 11 13:30:43 PDT 2009
What is the point to all this? Looks much more complex than I sketch
below, but I do not see value in the extra effort:
For 10 candidates, fill a 10x10 matrix - for A vs B need an entry
counting A>B and one for B>A.
With luck about 9 comparisons will identify winner since each
identifies a non-winner.
If A>B = B>A, keep both for the moment. If they lose to another, that
ends them. If all others lose to them, we have a tie.
Could be cycles. Comparing winner against 8 losers should identify
innocence vs guilt.
Dave Ketchum
On Oct 11, 2009, at 12:00 PM, Michael Rouse wrote:
> 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
More information about the Election-Methods
mailing list