Presumably each purchase is independent and can buy each card with equal probability $\frac{1}{30}$.
Suppose $P(n,k)$ is the probability that after $n$ purchases you are missing $k$ lucky cards. Then $P(n+1,k)=\frac{30-k}{30}P(n,k)+\frac{k+1}{30}P(n,k+1)$ starting at $P(0,6)=1$ and $P(0,k)=0$ for $k \not = 6$. This can easily be calculated, for example using a spreadsheet.
You are interested in the smallest $n$ such that $P(n,0) \gt 0.8$. It turns out that $P(97,0) \approx 0.79402$ while $P(98,0) \approx 0.80031$.
As a check on the calculations, the expected number of cards needed to be bought to get all the lucky cards is both $\displaystyle\sum_{n=0}^\infty (1-P(n,0)) = 73.5$ and $30\left(\frac16 + \frac15 + \frac14 + \frac13 + \frac12 + \frac11\right)=73.5$.
Added
If purchases were without replacement (meaning that at $30$ purchases you have all the cards, lucky and unlucky), the recurrence becomes
$P(n+1,k)=\frac{30-n-k}{30-n}P(n,k)+\frac{k+1}{30-n}P(n,k+1).$
You now find $P(29,0) = 0.8$. This is not a surprise as the probability the final card purchased is lucky is $\frac{6}{30}=0.2$.