If I reset after each run, how many times do I need to repeat an experiment to have a 75% chance of running it on all the subjects? Suppose there is a bag with 100 marbles. And I can draw 5 marbles in one attempt. But after that draw I have to put the marbles back. In how many attempts do I have a 75% chance that I have drawn each marble at least once? 
I wrote a brute force program that calculated this: 
Marbles = 100. DRAW 5 at a time. Performing 100000 runs.
Runs completed = 10000
Runs completed = 20000
Runs completed = 30000
Runs completed = 40000
Runs completed = 50000
Runs completed = 60000
Runs completed = 70000
Runs completed = 80000
Runs completed = 90000
Runs completed = 100000
For 75% confidence you need 115 draws.

But how can we mathematically arrive at the same answer? 
 A: I am not sure if the following argument does not have a flaw (dependence/independence worries mainly at the last step) but here is the idea: 


*

*When you perform a single draw of 5 marbles the probability that a fixed marble A shows up is:  $ p = {99 \choose 4} / {100 \choose 5} = 5/100$. Why? Well, we have $100 \choose 5$ 5-element subsets in total. Marble A is a member of $99 \choose 4$ of these subsets. Hence this probability value.  

*So when you perform a single draw of 5 marbles the probability that marble A does not show up is:  $ 1-p = 95 / 100$ 

*Different draws are independent obviously. So the after N draws the probability that marble A does not show up is: $(1-p)^N$

*$\Rightarrow$ After N draws the probability that marble A shows up is: $1 - (1-p)^N$ 

*This last step I am not quite sure about... because I am not sure if I can multiply these probabilities (not sure if these events here are independent, my intuition tells me they are but somehow I am not fully convinced)... Anyway, the argument here goes like this. So now since we have 100 marbles (not just marble A) the probability that after N draws all of them show up is simply the product: $p_N = (1 - (1-p)^N)^{100}$ 
I put this formula into a computer program and I see that:
$p_N \ge 0.75$ when $N \ge 115$.  
How do we find the value $115$ mathematically? Well, by solving this inequality  
$(1 - (1-p)^N)^{100} \ge 0.75$  for $N$.  
We solve it and we get:  
$N \ge log_{\frac{95}{100}}(1 - 0.75^{0.01}) \approx 114.1 $ 
Since N is an integer, we get $N \ge 115 $ 
I was able to empirically confirm this numeric result.
So now I am slightly more convinced that my argument in the last step is OK.     
