Dear Jobst, After submitting it I realized that my most recent proposal (repeated below) is not monotone. If I am not mistaken, a change in the definition of "friend" would fix it, but then the word "ally" would be more appropriate.
New Definition: Two ballots (that indicate both favorite and also approved) are allied iff they share a candidate in the "also approved" category (whether or not they agree on favorite). To fix the first description of the method, replace the word "friends" with the word "allied" . I hope that To fix the second description, replace "a friend" with "an ally" . I hope that works. Forest ----- Original Message ----- From: Date: Sunday, July 6, 2008 3:59 pm Subject: Fun with Friends and Dice To: Jobst Heitzig , Cc: [email protected], > Dear Jobst, > > Your ingenious use of coins was very inspiring to me. It > encouraged me to come up with a dice throwing > realization of our benchmark function f(x) = 1/(5-4x) in another > solution of our challenge problem. > > Also, since our goal is mutually beneficial cooperation, let me > define two ballots to be "friends" of each other > iff they co-approve one or more candidates. > > First, I give the method without the benefit of the dice: > > 1. Draw a ballot at random. Let Y be its favorite, let Z be the > most approved of its approved candidates, and > let x be the percentage of ballots that are friends with this one. > > 2. Elect Z with probability f(x), else Y. > > Now here's the dice rolling version: > > 1. Draw a ballot at random. Let Y be its favorite, and let Z be > the most approved of its approved candidates. > > 2. Roll a die until some number k other than six shows on top. > If k = 1, then elect Z, else ... > > 3. Draw a new ballot at random. If this new ballot is a friend > of the first ballot, go back to step 2, else ... > > 4. Elect Y. > > This method, like yours, guarantees a probability proportional > to faction size for those factions that choose > to bullet, yet it gently encourages friendship. > > What do you think? > > Forest > > > ----- Original Message ----- > From: Jobst Heitzig > Date: Friday, July 4, 2008 9:45 am > Subject: Re: [Election-Methods] Challenge Problem > To: [EMAIL PROTECTED] > Cc: [email protected] > > > Hi again. > > > > There is still another slight improvement which might be > useful > > in > > practice: Instead of using the function 1/(5-4x), use the function > > (1 + 3x + 3x^7 + x^8) / 8. > > This is only slightly smaller than 1/(5-4x) and has the same > > value of 1 > > and slope of 4 for x=1. Therefore, it still encourages > unanimous > > cooperation in our benchmark situation > > 50: A(1) > C(gamma) > B(0) > > 50: B(1) > C(gamma) > A(0) > > whenever gamma > (1+1/(1+(slope at x=1)))/2 = 0.6, just as the > > other > > methods did. > > > > The advantage of using (1 + 3x + 3x^7 + x^8) / 8 is that then > > there is a > > procedure in which you don't need any calculator or random > > number > > generator, only three coins: > ---- Election-Methods mailing list - see http://electorama.com/em for list info
