Let $P(n,k)$ be the probability that the sum of $k$ $n$-sided dice numbered $1$ to $n$ is prime. It is trivial that
$$ P(n,1)=\frac{\pi(n)}{n} $$
With a little more effort, one can obtain the relation
$$ P(n,2)=\frac{1}{n^2}\left(\sum_{p\leq n}p - \pi(n)+(2n+1)(\pi(2n)-\pi(n))-\sum_{n<p<2n}p\right) $$
The case for $k=3$ is not so straightforwardly reducible.
I propose the following conjecture:
Given $n\in \mathbb{N}$, there are infinitely many $k$ such that $$P(n,k)\geq 0.5$$
For $n=1$, for every prime $p_i$ we have $P(1,p_i)=1$. So it is true for $n=1$.
The case for $n=6$:
$$P(6,1)=.5$$ $$P(6,2)=.42...$$ $$P(6,3)=.34...$$
Can someone give a bound on $P(n,k)$? I remember dabbling with this problem while an undergraduate at university. It recently came back to mind and I figured I'd share the fun.