[Election-Methods] Fun with Friends and Dice

[fsimmons at pcc.edu]

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: