[Election-Methods] Fun with Friends and Dice

[fsimmons at pcc.edu]

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: election-methods at lists.electorama.com,

> 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: fsimmons at pcc.edu
> Cc: election-methods at lists.electorama.com
> 
> > 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:
>