How many (hypothetical) packs do I need to get every card in a set?

Let $s$ be the number of cards. Buying $k$ cans of $m$ distinct cards can be done in

$$\binom{s}{m}^k$$

ways.

To count how many of these that fails to contain all cards, count first those that miss card $1$. Next count those that miss card $2$ but contain $1$. Then those that miss $3$ but contain $1$ and $2$, etc... Adding these counts up gives the total amount of possible unsuccessful draws.

The formula for this uses inclusion/exclusion (inner sum). The index $i$ correspond to counting those without card $i+1$

$$\sum _{i=0}^{s-1} \sum _{j=0}^i (-1)^j \binom{i}{j} \binom{s-1-j}{m}^k$$

$j=0$ counts how many cases there are with card $i+1$ missing. $j=1$ subtract those cases where, in addition, one or more of the cards that must be contained is/are missing. $j=2$ adds those cases where, in addition, two or more of the cards that must be contained is/are missing, etc...

In your case $k=66$ happens to be the smallest value such that $5\%$ or less cases are unsuccessful.