How many marbles do I need to blindly pick until I'm ~80-90% confident that I've extracted 1 marble of each color from my bag Here's my problem written in terms of a marble bag problem:

I have a bag of $90$ marbles and $30$ unique colors. Let's assume each
color is represented equally, so there are $3$ of each color marble. I
will pick one marble out of the bag at a time and not replace it for
$N$ picks. How many picks from the bag do I need in order to have a
good probability ($\text~0.8-0.9$ maybe) to have at least $1$ of each color?

In other words, if I have this bag of $90$ marbles with $30$ unique colors, imagine I want to take out marbles one by one and put them in separate boxes, but I have a blindfold on. I want to continue doing this only as long as I have to until I'm confident enough that I've extracted at least $1$ of each different color marble.
I do not have much background in mathematics, and after looking at other probability examples online and trying to answer this myself, I'm very stuck and in over my head. I've tried looking up similar problems, "how to array a pooled library", and got nothing. I believe I know how to calculate this for something small, like $3$ colors and $6$ picks, by determining valid combinations, permutations, and individual probabilities to add all together. But I need to scale it up considerably and don't know how. I have a feeling there's some shortcut or series to use, but I don't know what it is. Can anyone help? Or perhaps point me in the right direction?
 A: Using the principle of inclusion exclusion, the probability that your sample contains every color is
$$
\frac{1}{\binom{90}N}\sum_{k=0}^{30}(-1)^{k}\binom{30}k\binom{90-3k}{N}
$$
The smallest $N$ for which this exceeds $0.8$ is $72$, and the smallest $N$ for which this exceeds $0.9$ is $76$.
A: I am unsure about the $80\%-90\%$ confidence part(since I do not know how to measure confidence exactly). However, I know how to be absolutely sure that you have drawn at least one marble of each color.
Consider the worst-case scenario: you draw a marble, then draw a marble of the same color, then draw a marble of the same color again. Now that color has been completely exhausted. Suppose this happens for every color you pick, for $29$ colors. You have currently drawn $87$ marbles, and have $29$ unique colors. Then, the only remaining marbles are of the same color, and all you have to do is draw one more to get all $30$ colors, for a total of $88$ marbles.
You may also be interested in the Pigeonhole Principle if you have not already discovered it. As simple as it seems, it can be used to solve Olympiad-level problems.
