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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