On May 3, 2008, at 11:22 , Jobst Heitzig wrote:
Dear Juho,this sounds nice -- the crucial point is that we'll have to analyse what strategic voters will vote under that method! Obviously, it makes no sense to the A voters to reverse their C>B preference since that would eliminate C instead of B and will result in B winning instead of C...Did you look deeper into the strategic implications yet?
The strategic implications of the STV based method were already discussed in other mails. The clone problem is a problem in many real life election environments.
STV is just one option. Also other methods can be used to eliminate the most unwanted candidates in a way that gives all voters a say in some proportionalish way. Here are two additional candidates for the challenge.
result(x) = sum { max { minmaxmargin(y) | y is ranked below x in v } | v in votes } result(x) = min { max { minmaxmargin(y) | y is ranked below x in v } | v in votes }
One quick example calculation to clarify the intended behaviour of the sum based method.
51: A>C>B 49: B>C>A The minmaxmargin results are: A=2 B=-2 C=-2.result(A) = -2*51-100*49 = -5002 (let's assume an imaginary candidate that is ranked last in all votes => -100)
result(B) = -100*51+2*49 = -5002 result(C) = -2*51+2*49 = -8 C wins.The min based function may lead to ties. The sum based function may also lead to ties, as above for the second and third position. One possible tie breaker would be to count the number of voters having the lowest minmaxmargin results (then the next lowest etc.). (that would make the result of A better than the result of B in both sum based and min based methods)
Sorry about the delay in answering the mails and writing short mails. I'm quite busy at the moment with other duties.
Juho
Yours, JobstP.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 theworst 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 byHow 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, JobstJuho 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 (andunfortunately that method is not monotonic): it is called AMP and isdefined 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 nameany 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 meanrating 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 forcooperation".3. For each option z, in order of descending approval score, do thefollowing: 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 strategicequilibrium is when all B-voters offer (A,C) and at least 49 of theA-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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