[EM] (Possibly) new method/request for voting paradoxes. :)
Michael Rouse
mrouse1 at mrouse.com
Mon Oct 12 05:59:57 PDT 2009
The additional complexity is to break cycles and come up with a complete
preference order. In one of the examples below,
5: A>B>C
3: B>C>A
4: C>A>B
the cycle was resolved as A>B>C.
I just wanted to show that it picked the Condorcet winner in the absence
of cycles. Like other Condorcet methods, breaking cycles is where
paradoxes arise, like IIA, but I'm also interested in the other
paradoxes it runs into. I'm also kind of curious how the rank orders
compare to that of other methods, like Kemeny-Young.
Michael Rouse
PS It's early on the West Coast. I probably shouldn't write after I just
wake up. :)
Dave Ketchum wrote:
> 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
>
>
>
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