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In my previous question titled "Conditional combinations of balls in bowls", is there a distribution to fit $k$ when $d \gg M$? I mean, when $d$ is so large, what is the distribution of the total number of balls?

$M$ can be a number between $4$ to $64$ and $d$ (number of bowls) is very large, about $10000$. Each bowl can have $0,1,2,\ldots,M$ balls in it with equal probability. What is the distribution of total number of balls $k$? $\Pr(k=k_0)=$?

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Is this homework? – user10525 Aug 17 '12 at 13:01
"...with equal probability" is helpful but ambiguous. Perhaps you could describe the process by which the bowl populations and $k$ are determined? – whuber Aug 17 '12 at 13:25
@whuber Seems to me that the OP is saying the probability is 1/(M+1) for each number of balls from 0 to M. – Michael Chernick Aug 17 '12 at 13:31
@Michael Yes, but that leaves a huge amount of latitude. Is $k$ fixed in advance? Are there other possible dependencies among the bowl populations? – whuber Aug 17 '12 at 13:33
@whuber K is a random variable, so it is not fixed. Although not stated I would assume that each bowl's contents is independent of the others and 0<=k<=dM is the constraint on k. – Michael Chernick Aug 17 '12 at 14:05

Using uppercase $K$ for a random variable and lowercase $m,d,k_0$ for constants:

The number of balls in a particular bowl has a discrete uniform distribution with mean $\frac{m}{2}$ and standard deviation $\sqrt{\frac{m^2+2m}{12}}$.

Add up a large number $d$ of these i.i.d. then the the distribution of sum $K$ with mean $\frac{md}{2}$ and standard deviation $\sqrt{\frac{(m^2+2m)d}{12}}$ can be approximated using the central limit theorem, remembering $K$ is discrete with integer spacing.

So $$\Pr(K=k_0) \approx \Phi \left( \frac{k_0 +\frac12 -\frac{md}{2}}{\sqrt{\frac{(m^2+2m)d}{12}}} \right)- \Phi \left( \frac{k_0-\frac12 -\frac{md}{2} }{\sqrt{\frac{(m^2+2m)d}{12}}} \right) \approx \phi \left( \frac{k_0-\frac{md}{2}}{\sqrt{\frac{(m^2+2m)d}{12}}} \right) $$ where $\Phi$ is the cumulative distribution function and $\phi$ is the probability density function of a standard normal distribution.

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