Dear Forest,
glad you find time to delve somewhat deeper into these questions. Looks
like a good idea to make the probability of using Random Ballot depend
on some "degree of cooperation".
Two notes as of now:
I guess you meant
RABMAC*doc^M + RB*(1-doc^M)
instead of
RB*doc^M + RABMAC*(1-doc^M),
right?
And I'm not so optimistic about monotonicity: Consider n voters with
these ballots:
n-3: A favourite > B also approved
2: B favourite
1: C favourite
With large n, RB elects A almost surely while RABMAC elects B almost
surely. When the last voters switches to approving A also, this is still
the case, but the "degree of cooperation" you defined will increase by
almost 1/n. So your method will move probability from the RB-winner A to
the RABMAC-winner B while monotonicity demands an increase in A's
probability. Details:
Before:
doc = ((n-3)²+2(n-1)+1)/n² = 1 - 4/n + 8/n²
P(A) = 0*doc^M + (n-3)/n*(1-doc^M)
After:
doc' = ((n-2)²+2(n-1))/n² = 1 - 2/n + 2/n²
P(A)' = 1/n*doc'^M + (n-3)/n*(1-doc^M)
For M=1:
P(A)'-P(A) = 1/n - 2/n² + 2/n³ + (1-3/n)*(-2/n+6/n²)
= -1/n + 10/n² - 16/n³
which is negative for n>10, right?
Maybe this can be fixed somehow be altering the definition of "degree of
cooperation"?
Yours, Jobst
[EMAIL PROTECTED] schrieb:
Dear Jobst,
I have a another solution to the challenge problem:
p: A>C>>B q: B>C>>A
Here's the method:
0. The ballots are approval with favorite indicated.
1. First the total approvals are counted and the candidates listed in
order of approval.
2. For each ballot B, let f(B) be the approval score (as a percentage
of the number of ballots cast) of the highest candidate on the list
approved by B.
3. Let doc (for "degree of cooperation") be the average value of f(B)
over all ballots.
4. Determine the winner with the lottery RB*doc^M +
RABMAC*(1-doc^M), i.e. decide whether to use Random Ballot or "Random
Approval Ballot Most Approved Candidate" on the basis of the lottery
(doc^M, 1-doc^M), where M will be specified below.
In the challenge problem, let's suppose that C is rated at R and S in
the respective factions, with R > p and S >q, and that the approvals
are
x: AC x': A only y: BC y': B only
with x + x' = p, and y + y' = q.
Then as long as x + y > max(p, q) we have
doc = (x')^2 + (x+y)^2 + (y')^2, and
Prob(A wins) = doc^M*x'+ (1-doc^M)*p, Prob(B wins) = doc^M*y' +
(1-doc^M)*q, Prob(C wins) = doc^M*(x+y).
For later use, note that doc=1 @ (x, y) = (p, q),
and that the partial derivative of doc w.r.t. x or y @ (p, q) is 2,
so that the partial of doc^M w.r.t. x or y @ (p, q) is simply 2*M.
The "utility" expectation for the first faction is
E1 = P(A wins) + R*P(C wins)
If we evaluate the partial derivative w.r.t. x of E1 at the full
cooperation point (x, y) = (p, q), we get
-1 - 2*M*p + R*(1 + 2*M) = 2*M*(R - p) - (1 - R)
which is greater than zero when M is sufficiently large, since R > p.
Similarly the partial derivative of E2 w.r.t. y at the same point is
2*M*(S - q) - (1 - S),
which is greater than zero when M is sufficiently large.
Therefore, local unilateral defection from full cooperation won't pay
if M is sufficiently large.
If I am not mistaken, the method is monotone and satisfies your
property about proportional probability for those factions that
steadfastly approve only their favorite.
On a technical note, if we replace the lottery (doc^M, 1-doc^M) for
deciding which kind of random ballot to use with the lottery (g(doc),
1 - g(doc)), where g(t)=1 - (1 - t)^(1/2), then we don't have to
worry about M.
This works because (no matter how large M) the slope of g(t)
eventually dominates the slope of t^M as t approaches 1.
Nevertheless, for the sake of simplicity I suggest using t^5 instead
of g(t).
My Best,
Forest
----
Election-Methods mailing list - see http://electorama.com/em for list info
