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

Fri May 2 14:36:33 PDT 2008

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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