[Election-Methods] method design challenge + new method AMP

[Juho]

How about using STV or some other proportional method to select the  
n-1 worst candidates and then elect the remaining one?

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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