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

[Jobst Heitzig]

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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