I have including the EM alias On Mon, Feb 9, 2009 at 11:55 PM, <fsimm...@pcc.edu> wrote: > I like this concept, but I'm not sure exactly how to combine it with the above > "tyranny free" method. I'm sure it is worth pursuing.
I was thinking: - Each voter marks a favourite and approved compromises 1) Find the approval winner 2) Place all ballots which approve that winner in a pile and assign it to the approval winner 3) Repeat the process excluding any ballots which have been assigned to a pile 4) When all the ballots are assigned to a pile, pick a ballot at random 5) The candidate associated with the pile the ballot was picked from is declared the winner Notes: Step 4 could just be pick a pile with each pile having a weighting equal to the number of ballots in the pile Piles smallers than a threshold could be excluded. In that case, the voters corresponding to those piles should be allowed to reassign their probability. For example, they could rank the candidates and their ballot would be assigned to the pile of the highest ranked candidate. A bullet voter would automatically create/be assigned to a pile for his candidate, since he hasn't approved any other candidate, so the ballot cannot be assigned to a different pile. The favourite marking doesn't actually have any effect in this method. However, it doesn't handle the generic case well though. 55: A(100) B(70) C(0) 45: A(0) B(70) C(100) If the 45 faction votes 5: B+C 40: C and the 55 faction compromises 55: A+B The results are Round 1 A: 55 B: 60 C: 40 B wins, B pile becomes 5: B+C 55: A+B Remaining 40: C Round 2 C wins, C pile becomes 40: C The result is B: 60% C: 40% By compromising, the 55 faction effectively transferred its weight from A to B, without the 45 faction having to respond in kind. A rule that would help would be that a ballot can only be placed in a pile for a candidate who isn't the favourite on the ballot, if ballots with that candidate as favourite represent less than half of the ballots in that pile. This would change the outcome: 5: B+C* 40: C* 55: A*+B The results are Round 1 A: 55 B: 60 C: 40 B wins, B pile becomes 5: B+C* 55: A*+B A* ballots are 91% of the ballots The A favourite ballots are more than 50% of the pile, so 50 of them are removed to balance the pile. B pile becomes 5: B+C* 5: A*+B A* ballots are 50% of the ballots Remaining 50: A*+B 40: C Round 2 50: A*+B 40: C A wins (B pile has already been created) A pile is 50: A* + B Round 3 40: C C wins C pile is 40: C Result A: 50% B: 10% C: 40% The 50% rule means that the 55 faction has effectively said that they will only move one ballot from A to B for every ballot the 45 faction moves from C to B. If both factions were to compromise, the result would be B wins round 1 and has pile 55: A*+B 45: B+C* 10 A*+B ballots are removed to make it 50% B pile 45: A*+B 45: B+C* Remainder 10: A*+B Remainder goes to A's pile. Result: 90%: B 10%: A This is reasonably fair, as it recognises the fact that faction A is larger. Also, it is an improvement over the previous non-compromise result from the perspective of both factions. ---- Election-Methods mailing list - see http://electorama.com/em for list info
