This seems to be a simple question but I can't seem to wrap my head around it.
let's say we throw $L$ marbles onto a $K*N$ grid, such that each spot in the grid can hold at most one marble. how many columns do we expect will have at least one marble on it?
If we throw 1 marble, exactly one column will be have a marble.
If we throw a second marble, there is $p=(K-1)/(KN-1)$ probability that the second marble land in the same column as the first one, and probability $1-p$ that the number of non-empty columns is 2.
I can see no way to get to $L$ marbles from here. I can't find a way to use the inclusion-exclusion principle cleverly enough to get an answer either.
This seems quite similar to Compute the expected number of tiles needed filled to fill a row in a grid, but I can't quite make the connection.
Expanding on @YJT's hint in the comments, the number (say $X$) of columns with at least one marble can be written as $X = X_1+\cdots + X_N$ where $X_i = 1$ if the $i^{th}$ column has at least one marble and $X_i=0$ otherwise.
Then $\mathbb{E}(X) = \mathbb{E}\left(\displaystyle\sum_{i=1}^{N}X_i\right) = \displaystyle\sum_{i=1}^{N}\mathbb{E}(X_i)$ as Expectation is linear.
Now $\mathbb{E}(X_i) = 1 \times P(\text{column i has at least 1 marble}) + 0 \times P(\text{column i has no marble})$ $\implies \mathbb{E}(X_i)= P(\text{column i has at least 1 marble})$
$P(\text{column i has at least 1 marble}) = 1 - P(\text{column i has no marble})=1-\dfrac{KN-K\choose L}{KN\choose L}$
So $\mathbb{E}(X) = N\left(1-\dfrac{KN-K\choose L}{KN\choose L}\right)$