Probability that all bins have at least 1 black ball when n black balls and N-n white balls are randomly partitioned into m equally sized bins?

Question

This post states the same problem I am looking at, with different variable names.
It does not look like the answer was accepted; and I have trouble following it, as it uses technical terms I am not familiar with (Hasse diagram, poset...).
I am trying to solve this using more elementary combinatorial principles, as a non-expert.

In essence, you have $N$ balls, $n$ of which are black, $N-n$ white.
You want to partition them into $m$ bins of size $N/m$, where $N/m$ is an integer.
The question is: what fraction of all possible partitions have at least one black ball in all bins?

Despite thinking about this quite hard, and sketching out several solutions based on my own reasoning and on people's feedback, e.g. by looking at occupancy theory and restricted partitioning, I consistently got stuck with summations of terms that I could not figure out how to obtain.

Would anyone be able to point me to literature or posts that describe this problem, if any exist, or suggest how to approach it?

Another question that would be useful to know the answer to is : can this be resolved by fairly elementary methods, or does it require particularly advanced ones?

Here are some pictures from my previous attempts.

All (20) possible sequences made from 3 black + 3 white balls:

Partitioned into 3 bins:

Partitioned into 2 bins:

I wonder if I am approaching this the wrong way and hence missing something obvious that would make it much easier to solve.

---

EDIT writing out the explicit formula resulting from @JMoravitz's answer, for the record

$$P(N,n,m) = \sum_{i=1}^m {(-1)^{i-1} \cdot \binom m i \cdot \frac {\binom {N-n} {\frac {i \cdot N} {m} }} { \binom {N} {\frac {i \cdot N} {m} } } }$$

$P(N,n,m)$ = probability that, by randomly partitioning $N$ balls, $n$ of which are black, $N-n$ white, into $m$ bins of equal integer sizes $N/m$, at least one bin contains only white balls.

And, if I am not mistaken, the rightmost factor:

$\frac {\binom {N-n} {\frac {i \cdot N} {m} }} { \binom {N} {\frac {i \cdot N} {m} } } = H(\frac {i \cdot N} {m}, N-n, n, \frac {i \cdot N} {m})$

i.e. the hypergeometric probability to draw $\frac {i \cdot N} {m}$ white balls from a set containing $N-n$ white balls and $n$ black balls, by randomly drawing $\frac {i \cdot N} {m}$ balls.

For full clarity, the answer to the original question of this post is $1-P(N,n,m)$ (probability that, by randomly partitioning $N$ balls, $n$ of which are black, $N-n$ white, into $m$ bins of equal integer sizes $N/m$, all bins contain at least one black ball.

It may be noted that when $n > 0$, it is sufficient to sum up to $m-1$, as the probability that all bins are empty is null.
In fact the upper limit of the summation could be decreased even further when $n > N/m$. The hypergeometric probabilities go to $0$ in those cases.

---

EDIT 2 adding R script, for convenience

N <- 6 # total size of set
n <- 3 # number of black balls in set
m <- 3 # number of equal-sized bins to make (must be a divisor of N)
P2 = sum(sapply(1:m, function(i) dhyper(i * N / m, N - n, n, i * N / m) * choose(m, i) * (-1)^(i - 1)))
print(P2)

---

EDIT 3 adding a more generic R script, handling any arbitrary set sizes

N <- 1000 # total size of set
n <- 10 # number of black balls in set
sizes <- c(100, 100, 800) # sizes of subsets to make (all >= 1, and must sum up to N)
P <- function(N, n, sizes) {
  if (!(sum(sizes) == N)) stop("The sum of sizes must be N.")
  if (any(sizes <= 0)) stop("No size can be <= 0.")
  if (any(sapply(sizes, function(x) floor(x) != x))) stop("Sizes must be integers.")
  M = length(sizes)
  sum(
    sapply(1:M,
           function(i) {
             (-1)^(i-1) *
               sum(
                 combn(sizes, i, 
                       function(s) dhyper(sum(s), N-n, n, sum(s))
                 )
               )
           }
    )
  )
}
print(P(N, n, sizes))
Function that calculates P for M subsets of sizes as close as possible to N/M
M <- 5 # number of subsets to make (must be <= N)
PE <- function(N, n, M) {
  if (M > N) stop("M cannot be greater than N.")
  if (M <= 0) stop("M must be positive.")
  if (floor(M) != M) stop("M must be an integer.")
  MC = N - floor(N/M) * M
  sizes = c(rep(ceiling(N/M), MC), rep(floor(N/M), M-MC))
  P(N, n, sizes)
}
print(PE(N, n, M))

Answer 1

Answer the opposite question of finding the probability that at least one bin has no black balls. (The "original question" apparently according to comments)

Let $N$ be the total number of balls, $n$ the number of balls which are black, and $k$ the number of bins these are being evenly split into (making $N$ a multiple of $k$) and an amount of balls in each bin being $N/k$.

Let $A_i$ be the event that the $i$'th bin has no black balls for $1\leq i\leq k$. We are trying to calculate $\Pr(\bigcup\limits_{i=1}^k A_i)$

This expands by inclusion-exclusion as

$\Pr(A_1)+\Pr(A_2)+\dots+\Pr(A_k)-\Pr(A_1\cap A_2)-\Pr(A_1\cap A_3)-\dots + \Pr(A_1\cap A_2\cap A_3)\pm\dots \pm \Pr(A_1\cap \dots \cap A_k)$

Investigating each of these individually we find that if we have the intersection of... say... $m$ of these events together, that corresponds to the event that those $m$ bins respectively all had no black balls. This occurs with probability $\dfrac{\binom{N-n}{mN/k}}{\binom{N}{mN/k}}$. That is, we need to choose an appropriate number of balls from the overall amount for these bins... and we compare this to the number of ways we had pulled from only the white balls.

Now, also recognize the high amount of symmetry, and that all terms in the expansion can be seen as equal to all others in the expansion involving the same number of events.

For your example of $N=6,n=3,k=3$ we get the calculation:

$$3\cdot \dfrac{\binom{3}{2}}{\binom{6}{2}} - 0+0 = \dfrac{3}{5}=0.6$$

Meaning the answer to the question of the question asked in your post is $1-0.6=0.4$, exactly as expected.