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 >>> >>> >>> >>> ___________________________________________________________ All >>> new Yahoo! 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