How long does it take to fill a bucket to capacity if randomly adding items to buckets? This is a toy problem I'm having trouble solving (not homework). I appreciate any help or pointers.
You are given $m=64$ buckets. Each 'run' you place a bean in a random bucket. After how many 'runs' (assigning a bean randomly to a bucket) do you have a 50% chance of filling a particular bucket to (or beyond) its capacity $c=8$?
 A: Let $X_i$ be the number of beans added to the bucket at run $n,$ $n=1,2,3,\ldots.$  The total number of beans in the bucket after run $n$ therefore is $Y_n= X_1+X_2+\cdots+X_n.$  Since $X_n$ has a Bernoulli$(1/64)$ distribution, $Y_n$ has a Binomial$(n,1/64)$ distribution.  Let $F_n(x) = \Pr(Y_n \le x)$ be its probability function.
The chance that the bucket is filled to capacity after run $n$ is 
$$\Pr(Y_n \ge 8) = 1 - \Pr(Y_n \le 7) = 1 - F_n(7).$$
We therefore need to find the first (smallest) $n\gt 0$ for which
$$1-F_n(7) \ge 50\% = \frac{1}{2}.$$

The search for this solution can be assisted by an estimate.  For instance, the distribution of $Y_n$ will be closely approximated by a Poisson$(n/64)$ distribution, whose median is close to $n/64.$  We might therefore search for solutions between $n_0 = 7(64)$ and $n_1 = 8(64).$  Indeed,
$$1-F_{7(64)}(7) \lt 0.40128 \lt \frac{1}{2} \lt 1-F_{8(64)}(7)\lt 0.54814,$$
suggesting $n$ is around $(0.548 - 0.5)/(0.548-0.401) = 0.33$ of the way between $7(64)$ and $8(64),$ or approximately $491.$  Indeed,
$$1 - F_{490}(7) = 0.49887 \lt \frac{1}{2} \lt 0.50114 = 1-F_{491}(7)$$
shows $n=491$ must be the answer.

Let's learn more by simulating the experiment.  Here's R code to run it 10,000 times, storing how many beans are needed each time:
n <- 1e4
set.seed(17)
Simulation <- replicate(n, 
  which.max(cumsum(rbinom(6*491, 1, 1/64)) >= 8)
)

A histogram of the number of runs indicate it is indeed centered close to $n=491:$

The average number of runs in this simulation was 511, ranging from 95 through 1490.  It helpfully shows us how variable the required number of runs is likely to be.
