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
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