[Election-Methods] [english 95%] Re: [english 94%] Re: method design challenge +new method AMP
Juho
juho4880 at yahoo.co.uk
Mon May 5 14:43:23 PDT 2008
On May 3, 2008, at 11:22 , Jobst Heitzig wrote:
> Dear Juho,
>
> this sounds nice -- the crucial point is that we'll have to analyse
> what strategic voters will vote under that method! Obviously, it
> makes no sense to the A voters to reverse their C>B preference
> since that would eliminate C instead of B and will result in B
> winning instead of C...
>
> Did you look deeper into the strategic implications yet?
The strategic implications of the STV based method were already
discussed in other mails. The clone problem is a problem in many real
life election environments.
STV is just one option. Also other methods can be used to eliminate
the most unwanted candidates in a way that gives all voters a say in
some proportionalish way. Here are two additional candidates for the
challenge.
result(x) = sum { max { minmaxmargin(y) | y is ranked below x in v }
| v in votes }
result(x) = min { max { minmaxmargin(y) | y is ranked below x in v }
| v in votes }
One quick example calculation to clarify the intended behaviour of
the sum based method.
51: A>C>B
49: B>C>A
The minmaxmargin results are: A=2 B=-2 C=-2.
result(A) = -2*51-100*49 = -5002 (let's assume an imaginary candidate
that is ranked last in all votes => -100)
result(B) = -100*51+2*49 = -5002
result(C) = -2*51+2*49 = -8
C wins.
The min based function may lead to ties. The sum based function may
also lead to ties, as above for the second and third position. One
possible tie breaker would be to count the number of voters having
the lowest minmaxmargin results (then the next lowest etc.). (that
would make the result of A better than the result of B in both sum
based and min based methods)
Sorry about the delay in answering the mails and writing short mails.
I'm quite busy at the moment with other duties.
Juho
>
> Yours, Jobst
>
>> P.S. It is quite easy to use also other methods than STV since
>> the combinatorics are not a problem. There are only n different
>> possible outcomes of the proportional method (if there are n
>> candidates). In this example it is enough to check which one of
>> the sets {A,B}, {A,C} and {B,C} gives best proportionality (when
>> looking at the worst candidates to be eliminated from the race).
>> Juho
>> On May 2, 2008, at 23:59 , Juho wrote:
>>> Here's an example on how the proposed method might work.
>>>
>>> I'll use your set of votes but only the rankings.
>>> 51: A>C>B
>>> 49: B>C>A
>>>
>>> Let's then reverse the votes to see who the voters don't like.
>>> 51: B>C>A
>>> 49: A>C>B
>>>
>>> Then we'll use STV (or some other proportional method) to select 2
>>> (=3-1) candidates. STV would elect B and A. B and A are thus the
>>> worst candidates (proportionally determined) that will be
>>> eliminated.
>>> Only C remains and is the winner.
>>>
>>> - I used only rankings => also worse than "52 point" compromise
>>> candidates would be elected
>>> - I didn't use any lotteries => C will be elected with certainty
>>>
>>> Juho
>>>
>>>
>>>
>>> On May 2, 2008, at 22:29 , Jobst Heitzig wrote:
>>>
>>>> Dear Juho,
>>>>
>>>> I'm not sure what you mean by
>>>>> How about using STV or some other proportional method to select
>>>>> the n-1 worst candidates and then elect the remaining one?
>>>> Could you give an example or show how this would work out in the
>>>> situation under consideration?
>>>>
>>>> Yours, Jobst
>>>>
>>>>> Juho
>>>>> On Apr 28, 2008, at 20:58 , Jobst Heitzig wrote:
>>>>>> Hello folks,
>>>>>>
>>>>>> over the last months I have again and again tried to find a
>>>>>> solution to
>>>>>> a seemingly simple problem:
>>>>>>
>>>>>> The Goal
>>>>>> ---------
>>>>>> Find a group decision method which will elect C with near
>>>>>> certainty in
>>>>>> the following situation:
>>>>>> - There are three options A,B,C
>>>>>> - There are 51 voters who prefer A to B, and 49 who prefer B
>>>>>> to A.
>>>>>> - All voters prefer C to a lottery in which their favourite
>>>>>> has 51%
>>>>>> probability and the other faction's favourite has 49%
>>>>>> probability.
>>>>>> - Both factions are strategic and may coordinate their voting
>>>>>> behaviour.
>>>>>>
>>>>>>
>>>>>> Those of you who like cardinal utilities may assume the
>>>>>> following:
>>>>>> 51: A 100 > C 52 > B 0
>>>>>> 49: B 100 > C 52 > A 0
>>>>>>
>>>>>> Note that Range Voting would meet the goal if the voters would be
>>>>>> assumed to vote honestly instead of strategically. With strategic
>>>>>> voters, however, Range Voting will elect A.
>>>>>>
>>>>>> As of now, I know of only one method that will solve the problem
>>>>>> (and
>>>>>> unfortunately that method is not monotonic): it is called AMP
>>>>>> and is
>>>>>> defined below.
>>>>>>
>>>>>>
>>>>>> *** So, I ask everyone to design some ***
>>>>>> *** method that meets the above goal! ***
>>>>>>
>>>>>>
>>>>>> Have fun,
>>>>>> Jobst
>>>>>>
>>>>>>
>>>>>> Method AMP (approval-seeded maximal pairings)
>>>>>> ---------------------------------------------
>>>>>>
>>>>>> Ballot:
>>>>>>
>>>>>> a) Each voter marks one option as her "favourite" option and may
>>>>>> name
>>>>>> any number of "offers". An "offer" is an (ordered) pair of
>>>>>> options
>>>>>> (y,z). by "offering" (y,z) the voter expresses that she is
>>>>>> willing to
>>>>>> transfer "her" share of the winning probability from her
>>>>>> favourite x to
>>>>>> the compromise z if a second voter transfers his share of the
>>>>>> winning
>>>>>> probability from his favourite y to this compromise z.
>>>>>> (Usually, a voter would agree to this if she prefers z to
>>>>>> tossing a
>>>>>> coin between her favourite and y).
>>>>>>
>>>>>> b) Alternatively, a voter may specify cardinal ratings for all
>>>>>> options.
>>>>>> Then the highest-rated option x is considered the voter's
>>>>>> "favourite",
>>>>>> and each option-pair (y,z) for with z is higher rated that the
>>>>>> mean
>>>>>> rating of x and y is considered an "offer" by this voter.
>>>>>>
>>>>>> c) As another, simpler alternative, a voter may name only a
>>>>>> "favourite"
>>>>>> option x and any number of "also approved" options. Then each
>>>>>> option-pair (y,z) for which z but not y is "also approved" is
>>>>>> considered
>>>>>> an "offer" by this voter.
>>>>>>
>>>>>>
>>>>>> Tally:
>>>>>>
>>>>>> 1. For each option z, the "approval score" of z is the number of
>>>>>> voters
>>>>>> who offered (y,z) with any y.
>>>>>>
>>>>>> 2. Start with an empty urn and by considering all voters "free
>>>>>> for
>>>>>> cooperation".
>>>>>>
>>>>>> 3. For each option z, in order of descending approval score,
>>>>>> do the
>>>>>> following:
>>>>>>
>>>>>> 3.1. Find the largest set of voters that can be divvied up into
>>>>>> disjoint
>>>>>> voter-pairs {v,w} such that v and w are still free for
>>>>>> cooperation, v
>>>>>> offered (y,z), and w offered (x,z), where x is v's favourite and
>>>>>> y is
>>>>>> w's favourite.
>>>>>>
>>>>>> 3.2. For each voter v in this largest set, put a ball labelled
>>>>>> with the
>>>>>> compromise option z in the urn and consider v no longer free for
>>>>>> cooperation.
>>>>>>
>>>>>> 4. For each voter who still remains free for cooperation after
>>>>>> this was
>>>>>> done for all options, put a ball labelled with the favourite
>>>>>> option of
>>>>>> that voter in the urn.
>>>>>>
>>>>>> 5. Finally, the winning option is determined by drawing a ball
>>>>>> from the
>>>>>> urn.
>>>>>>
>>>>>> (In rare cases, some tiebreaker may be needed in step 3 or 3.1.)
>>>>>>
>>>>>>
>>>>>> Why this meets the goal: In the described situation, the only
>>>>>> strategic
>>>>>> equilibrium is when all B-voters offer (A,C) and at least 49
>>>>>> of the
>>>>>> A-voters "offer" (B,C). As a result, AMP will elect C with 98%
>>>>>> probability, and A with 2% probability.
>>>>>>
>>>>>>
>>>>>>
>>>>>> ----
>>>>>> Election-Methods mailing list - see http://electorama.com/em for
>>>>>> list info
>>>>>
>>>>>
>>>>>
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>>
>>
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