Stochastic Urn Process using a Pareto distribution N urns are assigned m balls in a stochastic process based on a Pareto distribution. The process is as follows:
X is a Pareto random variable (xminimum = 1, alpha is a parameter)
if X > N, throw the ball out
Otherwise, put the ball in urn number floor(X)
Do this until m balls have been put in urns.  Then rank the urns by number of balls.
What should the final distribution look like, as N and m approach infinity? Does it resemble any standard prob. distribution? Doing Monte Carlo simulations, it seems to be heavy tailed, but not a Pareto distribution.  But I'm new to this - just getting my feet wet with stochastic processes.
Edit: Primary question is: What is prob. mass. function for balls in k-th urn (after ordering them).  But I'm open to answers along the lines of: Look at it differently.  Basic goal is - given a selection of choices, with balls assigned to them, but some choices a priori more popular, what is relationship between stochastic process and pmf of results?
(And, yes, if you can't tell, I'm coming from an applied perspective... this question is a simplification of a model encountered assigning awards to favorites).
 A: Ignoring the reranking (as jorki says, it seems to makes things too complicated), I think you can say something like:  
The probability that the any particular ball which goes into an urn turns out to go into the $k$th urn [$1 \le k < N $] is $$P_k = \frac{1/k^{\alpha} - 1/(k+1)^{\alpha}}{1-1/N^{\alpha}}$$ 
With $m$ balls in total in the urns, the probability that there are $m_k$ in the $k$th urn is 
$$\Pr (M_k=m_k) = \binom{m}{m_k} P_k^{m_k} (1-P_k)^{m-m_k}$$
i.e. a binomial distribution $B(m,P_k)$
Taking the distribution over the urns together, rather than individual urns, this is a multinomial distribution, with the parameters $m$ and $P_1, P_2, \ldots$
A: Got it! Unbelievable.  The process of accumulating and then ranking is identical to simply computing the CDF, then reflecting over the y=x axis (taking the inverse)!!!  P(Urn X has x balls) = M*Pareto(X).  Let f(z) = $CDF^{-1}(M*Pareto(z))$.  That means that (using frequentist terminology) n*z urns have less than f(z) balls.
So the final answer - to the original question without simplifying - is $n * CDF^{-1}(M*Pareto(x))$.  (Or, to preserve the order I mentioned in the question, reflecting that over the y-axis).
I'm still a novice here, so can I ask you experts to verify or disprove this?
