Say I have a bag of 100 unique marbles. With I replacement, I pick 10 marbles at a time, at random. Say I have a bag of 100 unique marbles. With replacement, I pick 10 marbles at a time, at random. How many times will I have to pick the marbles (10 marbles a pick) in order to have a 95% chance of having seen every unique marble at least once.

Simulations show that you need 73 or 72 picks (probably this, $10^7$ turns give $p = 0.950612$) and I don't know any efficient method to calculate that exactly (of course, that doesn't mean there isn't one). 
irb(main):021:0> average(1000000) do g(100,10,73)? 1:0 end
=> 0.955118
irb(main):022:0> average(1000000) do g(100,10,72)? 1:0 end
=> 0.950215
irb(main):023:0> average(1000000) do g(100,10,71)? 1:0 end
=> 0.945348

If you want some formulas, then very rough estimate using Markov's inequality gives you $\frac{49.9}{1-0.95} \simeq 998$ times. With such a difference, you could as well assume that you replace each marble after picking it (that is you pick a marble and then replace it, and again a marble, and again replace it), and then this is called the coupon collector's problem.
The expected time to see all unique marbles is $n\mathcal{H}_n$, where $n$ is the number of marbles and $\mathcal{H}_n = \sum_{k=1}^{n} \frac{1}{k}$ is the $n$-th harmonic number. Using this estimate you can use the Markov's inequality 
$$P(X \geq t) \leq \frac{\mathbb{E}X}{t}$$
to bound the necessary number of picks. With $n = 100$ we get
$$P(X \geq t) \leq \frac{100\mathcal{H}_{100}}{t}\leq5\%$$
that is, $0.05t \geq 100 \mathcal{H}_{100} \approx 518.737$, so $t \geq 10375$. Picking 10 at a time, you will have to pick 1038 times.
I hope this helps ;-)
