[EM] (Possibly) new method/request for voting paradoxes. :)
Dave Ketchum
davek at clarityconnect.com
Mon Oct 12 11:22:32 PDT 2009
To look for a better sway to resolve cycles is worthy. However, I
still do not see the gain in what you offer, considering the expense.
Each of the members of a cycle would be winner if only one of them ran.
When they are close to a tie it matters little which wins.
When far away the deciding is easy.
We have many competing methods already - choosing among them
needs more effort than has happened,
Dave Ketchum
On Oct 12, 2009, at 8:59 AM, Michael Rouse wrote:
> 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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