What is the probability of having no mutual ball in several draws? This question is related to this previous one.
Suppose we have 100 balls numbered from 1 to 100 in a bag.
Let's say I will make $n$ random draws with replacement of $x$ balls and let's note $\mathcal{S}_i$ the $x$ values of the $i^{\text{th}}$ draw.
At each draw, I only consider the balls I have already seen, i.e. $S_{i+1} = S_{i+1} \cap S_i$ (except for $S_1$).
My question is: given a fixed value for $x$ (let's say 14), how can I bound $n$ to have $\mathcal{S}_n = \emptyset$ with a high probability?
The answers for my previous question show that for two draws, the probability that $\mathcal{S}_2 = \emptyset$ (with $x= 14$) is $\frac{86}{100} \times \frac{85}{99} \times ... \times \frac{73}{87} \approx 0.10$. For which values of $n$ could I reach something like $\mathcal{S}_n > 0.95$?
 A: First, we can calculate the exact probability using inclusion-exclusion. Denote the number of balls by $m$, so in your case $m=100$. The probability that $k$ particular balls haven't been drawn yet after $n-1$ draws is
$$
\left(\frac{\binom{m-k}x}{\binom mx}\right)^{n-1}\;.
$$
Thus by inclusion-exclusion the probability $p$ that all $x$ balls drawn in the $n$-th draw had already been drawn in the previous $n-1$ draws is
$$
p=\sum_{k=0}^x(-1)^k\binom xk\left(\frac{\binom{m-k}x}{\binom mx}\right)^{n-1}\;.
$$
Second, since you want high probability to have already drawn all $x$ balls drawn, it will be unlikely that there's more than one undrawn ball, so we should get a good estimate for the complement of the desired probability from the expected number of undrawn balls:
\begin{align}
p&\approx1-x\left(\frac{\binom{m-1}x}{\binom mx}\right)^{n-1}\\
&=1-x\left(1-\frac xm\right)^{n-1}\;,
\end{align}
which is just the first term of the inclusion-exclusion sum. Solving for $n$ yields
$$
n\approx\frac{\log\frac{1-p}x}{\log\left(1-\frac xm\right)}+1\;.
$$
In your example, with $m=100$, $x=14$ and $p=0.95$, this yields
\begin{align}
n&\approx\frac{\log\frac{0.05}{14}}{\log\left(1-\frac{14}{100}\right)}+1\\&\approx38.36\;,
\end{align}
and plugging the next higher integer $n=39$ into the exact expression for the probability yields
$$
\sum_{k=0}^{14}(-1)^k\binom{14}k\left(\frac{\binom{100-k}{14}}{\binom{100}{14}}\right)^{39-1}\approx95.5\%
$$
as desired.
