how to calculate the odds (with repetition) Here's the case,
There are $8$ different items with equal chance to draw each of them. Items are put back between each draw.
How do we calculate the odds of picking all $8$ of them in $X$ draws?
I believe for $8$ draws, the odds can be calculated as $\dfrac{7!}{8^7} = 0.24\%$. What about for $9$ or more draws?
Is there also a way to find out how many draws are required to reach $X\%$ probability?
 A: You are correct, that $\frac{7!}{8^7}$ is the correct answer.  I will call it$\frac{8!}{8^8}$ for reasons that will be a little more clear.
Let's look at the X=9 case.  We would still have in the numerator 8! because that will still give us the correct structure we need.  However, there is an additional event here, that of which we pick incorrectly one of the times.  That incorrect pick can only come on the 2nd, 3rd, 4th, 5th, 6th, 7th or 8th pick, since we have probability 1 for the first pick and probability $\frac{1}{8}$ for the second.
Now, if we pick incorrectly on the 2nd, our probability of it being not a color we need is only $\frac{1}{8}$ since the only way that can occur is if we choose the same color as our first pick.  However, if we pick incorrectly only on the 3rd trial our probability is $\frac{2}{8}$ because now there are 2 colors we can choose that are not the colors we need.  Thus, for X=9, we have
$$P(X=9)=\frac{8!}{8^9}\sum_{i=1}^{7}{i}=.0084$$
It's a little higher, and we should expect this I think.
Now for X=10, we have the added issue that there is another trial to err at.  Thus, we can have errors at (2,3),(2,4),...,(8,8),(8,9) so there are 28 different combinations since $i\ge{j}$ (we can't have an error at (3,2) since it is the same as (2,3) and we already counted that), of having those errors throughout the 10 trials.  Thus, for X=10, we have
$$P(X=10)=\frac{8!}{8^{10}}\sum_{j=1}^{7}{\sum_{i=1}^{j}{ij}}=.01735$$
Then
$$P(X=11)=\frac{8!}{8^{11}}\sum_{k=1}^{7}{\sum_{j=1}^{k}{\sum_{i=1}^{j}{ijk}}}=.0276$$
I went ahead and calculated with Maple, the next $x=12,13,14$.  THus, $$P(X=k),k=12,13,14, = .03475, .04601, .0524.$$
You could also now say that $P(X\le{14})\approx{.192}.$
I think this should give the answer you are looking for.  As for finding the number of trials, I'm not sure.
