Let's say we have a 20-sided die and a 6-sided die. In order to roll the 6-sided die one time, we must first roll a 20 on the 20-sided die. Our objective is to get to the 6-sided die and roll each number once
How many rolls should it take us to roll each number on the 6-sided die? To me, it seems to be an extension of the https://en.wikipedia.org/wiki/Coupon_collector%27s_problem
My initial hypothesis is something like (20)(6/(6-0)) + (20)(6/(6-1)) + ... (20)*(6/(6-5))
AKA SUM i=0 to n-1 with (20)*(n/(n-i)), where n is subbed in for 6
Looking good? I don't really have anywhere to validate my results, as this is for a personal project
From the basic coupon collector's problem, you need an expected $\sum\limits_{i=1}^n \frac{n}{i}$ attempts to get the full set.
So here, with $n=6$, you need an expected $14.7$ rolls of the six-sided die.
For each roll of the six-sided die, you need an expected $20$ rolls of the twenty-sided die to see a $20$, using the geometric distribution. So in this case you need an expected $20\times 14.7=294$ rolls of the twenty sided die.
So overall you need an expected $14.7+294=308.7$ dice rolls.