[Election-Methods] [english 95%] Re: [english 94%] Re: method design challenge +new method AMP

Sat May 3 01:22:14 PDT 2008

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?

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.
>>>>>
>>>>>
>>>>>
>>>>> ----
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>>>>> list info
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