Rolling infinitely many dice For every integer $n>1$, I have a die $D_n$ with $a_n$ sides labeled $1$ to $a_n$.
If $a_n=n^k$, for integer $k>0$, and I roll all the dice at once, what is the probability none of them lands on the side labeled $1$? What is the probability exactly one lands on the side labeled $1$?

First, suppose there are total of $q$ dice.
Given that, for fixed $m$, $D_m = 1$, the probability that no other die rolled at one is:
$$
   \mathbb{P}(\forall_{n \not= m} \, {D_n \not= 1} ; D_m = 1 ) = \frac{1}{1-\frac{1}{a_m}} \prod_{n=2}^q \left( 1- \frac{1}{a_n} \right)
$$
The probability that exactly one die lands on 1 is:
$$
  p_k(q) = \sum_{m=2}^q \frac{1}{a_m} \mathbb{P}(\forall_{n \not= m} \, {D_n \not= 1} ; D_m = 1 ) = \prod_{n=2}^q \left( 1- \frac{1}{a_n} \right) \sum_{m=2}^q \frac{1}{a_m-1} = 
   \prod_{n=2}^q \left( 1- \frac{1}{n^k} \right) \cdot \sum_{m=2}^q \frac{1}{m^k-1}
$$
For small values of $k$ we have
$$
  p_1(q) = \frac{1}{q} H_{q-1} \qquad
  p_2(q) = \frac{3}{8} \left(1+\frac{1}{12 q}\right)\left(1-\frac{1}{q} \right)
$$
For higher order the probability is expressible in terms of $\Gamma$ functions and poly-gamma functions.
In the limit of infinitely many dice:
$$
   \lim_{q \to \infty} p_1(q) = 0 \qquad 
   \lim_{q \to \infty} p_2(q) = \frac{3}{8} \qquad
   \lim_{q \to \infty} p_3(q) = \frac{\cosh \left(\frac{\sqrt{3} \pi }{2}\right)}{9 \pi }
$$
