If I randomly throw 10,000 balls to 4,000 buckets, what is the probability that at least one bucket contains more than 10 balls? If I randomly throw 10,000 balls to 4,000 buckets, what is the probability that at least one bucket contains more than 10 balls?
I don't know how to even approach this.
A rough number is also accepted - it doesn't have to be precise.
 A: You have a Poisson distribution with mean $2.5$ (the average number of balls per bucket).  Find the chance that one bucket has at most $10$ balls, raise to the  $4000$ power to have them all have ten or less, and subtract from $1$.  This is not exact-if one bucket is low that increases marginally the chance that another is high, but it will be very close.
A: Let $X=(x_1, x_2 \cdots x_N)$ ($N=4000$) be the number of balls in each bucket; then $p_X(X)$ follows a multinomial distribution, with $\sum x_i = M$ ($M=10000$). Now, let $Y$ be $N$ iid Poisson variables with mean $\lambda$ -still not specified. Then the distribution of $Y$ conditioned on $\sum y_i = M$ is the same as our multinomial. Let $E$ be the event that all buckets have less than $K=10$ balls. Then
$$P(E) = P(y_i < K \mid \sum y_i = M)= \frac{P( \sum y_i = M \mid y_i < K )}{P(\sum y_i = M)} P(y_i<K)$$
where $y_i < K $ implies $\forall i$. 
Each factor can be computed or approximated :
$$P(y_i<K)= \left[ g(K,\lambda) \right]^N$$ 
where  $$g(K,\lambda) = e^{-\lambda}  \sum_{x=0}^{K-1} \frac{\lambda^x}{x!}=\frac{\Gamma(K,\lambda)}{\Gamma(K)}$$
Now, because the sum of iid Poisson variables is Poisson:
$$P(\sum y_i =M) = e^{-N\lambda} \frac{(N \lambda)^M}{M!}$$
The remaining factor  can be approximated using CLT as 
$$P( \sum y_i = M \mid y_i < K ) \approx \frac{1}{\sqrt{2 \pi N \sigma^2}}\exp{\left(-\frac{M-N\mu}{2N \sigma^2}\right)}$$
where $\mu$,$\sigma^2$ are the mean and variance of a Poisson with parameter $\lambda$ 
truncated to $0,K-1$.
A good selection of $\lambda$ would be that that gives $\mu N=M$, because that would probably produce a good approximation. 
