Problem on EU commission Consider the following problem.
A collection of $n$ countries $C_1, \dots, C_n$ sit on an EU commission. Each country $C_i$ is assigned a voting weight $c_i$. A resolution passes if it has the support of a proportion of the panel of at least $A$, taking into account voting weights. Each country $C_i$ has a probability $p_i$ of voting for the resolution, and each country acts independently of the others.
The problem is to assign the voting weights so as to maximize the probability that any given resolution will pass. I am interested in answering the question asymptotically under something like the following assumptions.


*

*The number of countries $n$ is very large. (Perhaps the EU's $n = 28$ is already not so far from this!)

*The proportion of votes held by any one country is bounded above by $M/n$, for some fixed reasonable number $M$.

*$p_i > A$ for all $i$.

*The probabilities $p_i$ are bounded away from $1$.
Perhaps some of these conditions can be relaxed, or perhaps additional assumptions are needed, but these are the ones that seem to be needed for my arguments below.
I have tried to answer the question in an approximate and non-rigorous way as follows. 
Let $X_i$ be the random variable equal to $1$ when country $C_i$ votes for the resolution, and $0$ otherwise. Now let $V = \sum c_i X_i$. By a suitably general version of the central limit theorem (the Berry-Esseen inequality?), $V$ follows approximately a normal distribution with mean $\sum c_i p_i$ and variance $\sum c_i^2 p_i(1-p_i)$. The probability that we would like to maximize is 
$$P\left( V \geq A\sum c_i \right).$$ 
If we let $F(z)$ be the cumulative distribution function for the standard normal distribution, this probability can be approximated by $F(z)$ where
$$z = \frac{\sum c_i(p_i - A)}{\left[\sum c_i^2 p_i (1-p_i) \right]^{1/2}}. $$
Considering the gradient of the function $z = z(c_1,\dots,c_n)$ shows that $z$ is maximal when the weights $c_i$ are proportional to the numbers
$$\gamma_i = \frac{p_i - A}{p_i(1-p_i)}.$$
I conclude that it is plausible that the weights $c_i = \gamma_i$ are close to being optimal.
My questions, in descending order of importance, are:


*

*Has anything significant been written on this problem, or an equivalent one?

*Is my "theorem" correct?

*What would a rigorous formulation of the "theorem" look like?
EDIT: I've simulated the problem for $A = 0.5$ with 1867 countries with a 50.17% chance of voting in favour and 637 countries with a probability of 50.5%. I gave weight $1$ to each of the first group of countries and weight $c$ to the second. In the graph below, the horizontal axis is for $c$, and the vertical axis for the probability of passing the resolution. The blue curve represents the theoretical probability we would have if the normal approximation worked perfectly, and the red curve experimental data based on 5 million repetitions of the experiment. The maximum for the red graph is not too far from the conjectured optimal value of $\gamma = 2.94$. 

EDIT: In response to a comment, here are some additional details on the maximization of $z$ above. By homogeneity, it makes no difference whether or not we constrain the $c_i$'s to have sum $1$. But if we do, then a compactness argument shows that $z$ must attain a maximum at some point. 
Now return to unconstrained $c_i$'s. $\partial z/\partial c_i$ has the same sign as
$$\frac{\sum c_j^2 p_j (1 - p_j)}{\sum c_j (p_j - A)} \gamma_i - c_i.$$
This shows that where the maximum occurs, all the $c_i$'s must be proportional to $\gamma_i$.
 A: This is a very interesting problem! On your theorem: I don't understand why you maximize $z$, which is just a density (up to that point everything seems fine). You want to maximize $P(V\ge A)$. You assume for sufficiently many countries that this can be approximated by a normal distribution, invoking some version of the CLT. Thus, 
$$P(V\ge A)\approx \int^\infty_A \phi\left(\frac{x-\sum c_ip_i}{\sqrt{\sum c_i^2 p_i(1-p_i})}\right) dx=1-\Phi\left(\frac{A-\sum c_ip_i}{\sqrt{\sum c_i^2 p_i(1-p_i)}}\right).$$
How to maximize this? I am not entirely sure. I remember that the normal distribution is log-concave, and the CDF of log-concave functions is log-concave. So if $log(\Phi(x))$ is concave, then $-log(\Phi(x))$ is convex. But that doesn't help us here.. 
A few more suggestions:


*

*On assumption 2): Why do you need $M$ if you defined weights $c_i$ already?

*In your formulation, you could constrain the weights to $\sum_i c_i=1$, then the condition is $V\ge A$, looks nicer but is not necessary.

*On assumption 3): I agree it doesn't make the problem meaningless, but it seems unnecessary - it doesn't change the maximization problem. It just implies that, say, even an equal voting weight distribution would lead to more passes than failures of resolutions. But the problem of $p_i<A$ for some (or even all) countries would still be interesting. Given the normal approximation, there is still a positive probability for the resolution to pass whenever $p_i>0$ for all $i$, but it would be harder. This way you could model "harder resultions".

*On assumption 1) and 3): if $n\to\infty$, then assumption 3 guarantees that the resolution passes, as long as you have positive weight $c_i$ on all countries. Because the expected voteshare is above $A$, and asymptotically the expected vote share realizes with probability 1 (some strong law of large numbers). Interesting: if some countries have $p_i>A$ (a positive mass, to be exact) and some $p_i<A$, then as $n\to\infty$ you can guarantee passing of the resolution by giving positive weights to all with $p_i>A$ and none to the others.

*Given the previous point, it seems dangerous to talk about "asymptotics" - you just want finitely but many countries so that you can approximate with a normal distribution, but you don't actually want $n\to\infty$ as the problem then is trivial given assumption 3). Maybe this is what the commenter above meant.

*Where can you find something similar? I think your best bet might be the finance literature, where you compute the probability that your portfolio investment return is above some threshold $A$. There are some assets which have a similar structure as these votes (e.g., bonds): either the asset pays a positive dividend or the issuer goes bankrupt and the return is zero - just like your random variable $X_i$. Same for a loan portfolio.

*Your comments about "getting a decision right" reminded me of the Condorcet jury theorems. The setting is a special case of yours, where every member has the same voting weight and the majority threshold is $A=1/2$

*It also reminded me of the Feddersen Pesendorfer game theoretic analysis of unanimity voting rules. Neither of the two are directly related to your problem though. Again, finance seems to be your best bet.

A: http://en.m.wikipedia.org/wiki/Roy's_safety-first_criterion
(About related finance problems).
trying to chose a portfolio that maximizes the probability of meeting a threshold is probably close to the problem you're mentioning here.
As for the EU, their goal when they set up the system was not too far from what you're describing. The idea was also to prevent the formation of coalitions against France and Germany (if those 2 agree on something, it's very difficult to block it, and inversely, it's very difficult to pass a law without them agreeing).
A: Forgive me if this solution may appear to be too "simplistic", maybe I misunderstood something in your original question, but what I gathered is that you have to set the various votes of countries in a manner that you maximize the probability of a resolution passing.
Strangely enough for all the assumptions you mentioned, 3 of them did not matter in my solution and only number 2 had a meaning.
I did add one additional assumption though. All countries are "sorted" by decreasing probability of passing the resolution, that means:
$$ p_n \le p_{n-1} \le ... \le p_2 \le p_1$$
It should not affect generality too much, since the probability is data (from what I understand).
Let w be the relative weight of votes:
$$ w_i = {c_i \over \sum c_i}$$
It follows that:
$$ \sum w_i = 1 $$
Also let m be the bound for the proportion of votes:
$$ m = {M \over n} $$
Consider that m is less than 1 so we can write the following:
$$ {1 \over m}  = k + \eta$$
Where k is an integer and eta is a real number higher than or equal to 0 but lower than 1.
This is useful for telling how many countries can have an assigned weight of m, and, in fact we're gonna assign m to exactly the K countries with the highest probability. (the first K)
$$ w_1 = ... = w_k = m $$
Country K + 1 will get the "leftover" (note that if eta is zero, this country gets zero votes):
$$ w_{k+1} = 1 - m \cdot k $$
Everyone else gets zero votes:
$$ w_{k+2} = ... = w_n = 0 $$
Note that this values are weights, to go from weights to actual votes you simply have to multiply all weights for the same arbitrary constant.
Again, since this appears to be too "simplistic", it's very likely that I missed something in your original question or misunderstood it entirely. If that's the case, please tell me and I will deleted this answer.
