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 $1$?
 A: 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:
$$
   \lim_{q \to \infty} p_1(q) = 0 \qquad 
   \lim_{q \to \infty} p_2(q) = \frac{3}{8} \qquad
   \lim_{q \to \infty} p_3(q) = \frac{\cosh \left(\frac{\sqrt{3} \pi }{2}\right)}{9 \pi }
$$
A: For each die, the probability of landing on the side labeled one is one over the number of faces of the die, and the probability of landing on a different face is one minus that. Each die is independent, so we can simply multiply the probabilities. The probability that none of them land on the face labeled one is then
$$
\prod_{n=2}^\infty 1-\frac{1}{n^k}
$$
I know no way of solving the general case, but we can have Mathematica calculate the result for various values of $k$, and the results are pretty messy.

