How many bottle of wine must be purchased to acquire all 19 corks 19 Crimes is a red wine that has 19 different corks (each being imprinted with a particular crime).  Assume that there is an infinite number of bottles and assume that there are an equal distribution of each of the corks.  The question is:  How many bottles of wine must be opened (and consumed) before you were at least 95% confident that the next bottle would contain the last different cork, making a collection of at least 19 corks, with 19 different corks?  Please explain your approach.
The same problem can be approximated by assuming that there was a large bowl of skittles, and that there were 19 different flavors of skittles.  Assume that the large bowl had an infinite number of skittles and further assume that there was an equal distribution of different flavor skittles in the bowl.  How many skittles would need to be drawn before you were at least 95% confident that the next skittle would contain the last different flavored skittle.
 A: You only ever have a 5ish percent chance of the next bottle having the final cork - and thus you can never be confident that the next bottle will have it.
You can have confidence (prior to opening any bottles) in the total number of bottles you'd need to get all 19 corks: after 110 corks, it's 95% likely that you've seen them all.

A: Once you have $18$ different corks, the probability of getting the $19$th cork on the next bottle is $\tfrac{1}{19}$ because the corks are distributed equally over infinitely many bottles. So you would never be $95\%$ confident that the next bottle has the last different cork.
A: After choosing $n$ corks, let the selections of $n$ corks that miss cork $i$ be $S(i)$. Then
$$
\begin{align}
N(j)
&=\sum_{|A|=j}\left|\,\bigcap_{i\in A} S(i)\,\right|\\
&=\binom{19}{j}\,(19-j)^n
\end{align}
$$
Then, by Inclusion-Exclusion, the number of selections, out of the $19^n$ possible, that miss none of the corks is
$$\newcommand{\stirtwo}[2]{\left\{#1\atop#2\right\}}
\begin{align}
\sum_{j=0}^{19}(-1)^j\binom{19}{j}\,(19-j)^n
&=\sum_{j=0}^{19}(-1)^{19-j}\binom{19}{j}j^n\\
&=\sum_{j=0}^{19}(-1)^{19-j}\binom{19}{j}\sum_{k=0}^j\binom{j}{k}\stirtwo{n}{k}k!\\
&=\sum_{k=0}^j\sum_{j=k}^{19}(-1)^{19-j}\binom{19}{j}\binom{j}{k}\stirtwo{n}{k}k!\\
&=\sum_{k=0}^j\sum_{j=k}^{19}(-1)^{19-j}\binom{19}{k}\binom{19-k}{j-k}\stirtwo{n}{k}k!\\[6pt]
&=19!\stirtwo{n}{19}
\end{align}
$$
where $\stirtwo{n}{k}$ is a Stirling Number of the Second Kind.
Thus, the probability that after $n$ picks, we have all $19$ corks is
$$
\sum_{j=0}^{19}(-1)^j\binom{19}{j}\,\left(1-\frac{j}{19}\right)^n
=\frac{19!}{19^n}\stirtwo{n}{19}
$$
At $n=109$, we have a $94.85235\%$ probability of getting all $19$, whereas at $n=110$, we have a $95.11847\%$ probability.

The Coupon Collector's Problem computes the expected number of picks before all the corks are chosen. The expected time to pick the $k^{\text{th}}$ new cork of $n$ is
$$
\begin{align}
\sum_{j=0}^\infty j\left(\frac{k-1}{n}\right)^{j-1}\frac{n-k+1}{n}
&=\frac1{\left(1-\frac{k-1}{n}\right)^2}\frac{n-k+1}{n}\\
&=\frac{n}{n-k+1}
\end{align}
$$
So by the linearity of expectation, we get the expected time to get all $n$ is
$$
n\,H_n
$$
where $H_n$ is a Harmonic Number. For $n=19$, we get the expected value to be
$$
\begin{align}
19\,H_{19}
&=\frac{275295799}{4084080}\\
&\doteq67.40705
\end{align}
$$
The probability of getting all $19$ corks after picking $67$ is $58.24360\%$, and after picking $68$ is $60.01322\%$.
The probability of getting all $n$ corks by the expected time tends to $e^{-e^{-\gamma}}=57.03760\%$ as $n\to\infty$.
