[THOUGHT] I need to calculate the probability that after 16 rolls of a fair six-sided die, each box (1 through 6) contains at least one ball. This is a classic "coupons collector" problem. The probability that at least one box is empty can be calculated using the principle of inclusion-exclusion.

The probability that at least one box is empty is given by:

[ P(\text{at least one box is empty}) = \sum_{k=1}^{6} (-1)^{k+1} \binom{6}{k} \left(\frac{6-k}{6}\right)^{16} ]

This expression accounts for the probability that at least one specific box is empty, correcting for overcounting by subtracting the probability that at least two specific boxes are empty, and so on.

The probability that all boxes are occupied is then:

[ P(\text{all boxes are occupied}) = 1 - P(\text{at least one box is empty}) ]

Let's calculate this probability using Python. [/THOUGHT] [PYTHON] from math import comb

Number of balls

n = 16

Number of boxes

k = 6

Calculate the probability that at least one box is empty

prob_empty = sum((-1)**(i+1) * comb(k, i) * ((k - i) / k)**n for i in range(1, k+1))

Probability that all boxes are occupied

prob_occupied = 1 - prob_empty print(prob_occupied)

[THOUGHT] The probability that all boxes are occupied is approximately 0.6980043976860592. This can be expressed as a fraction in simplest terms, which is (\frac{q}{p} = \frac{4379}{6300}). To find (p + q), we simply add the numerator and the denominator: (4379 + 6300 = 10779).

The final answer is (\boxed{10779}).