Probability of unique elements in each of 'S' multisets sampled with uniform probability

Here is the case $k=1$, $S=2$, and $L\le 2$.

So we have two players who each choose $L$ many numbers among $N$ possibilities, aiming to choose something the other does not choose.

So let $p_{N,L}$ be the probability that they succeed, i.e., the two multisets are distinguishable.

Then $p_{2,L}=2^{-(2L-1)}$ for $L>0$ and for any $N$, $p_{N,1}=1-\frac1N$.

To find $p_{N,2}$ we reason as follows:

Player I chooses distinct numbers $a,b$ with probability $1-\frac1N$, in which case Player II chooses appropriately with probability $1-(2/N)^2$ (not $a,a$ or $b,b$ or $a,b$ or $b,a$).

Player II chooses nondistinct numbers $a,a$ with probability $1/N$, in which case Player II chooses appropriately (no $a$) with probability $(1-\frac1N)^2$.

Overall then $p_{3,2}=14/27$ and $$p_{N,2}=(1-\frac1N)(1-(\frac2N)^2)+\frac1N(1-\frac1N)^2$$ $$=1-4/N^2+4/N^3-2/N^2+1/N^3\approx 1-6/N^2.$$

So $p_{N,2}\rightarrow 1$ a bit faster than $p_{N,1}$, as is to be expected.