Expected size of collection based on number of uniques I have been thinking about this for some time. Imagine there are fortune cookies. Each of them can carry a random quote (1 out of total 100 possible quotes). After buying N cookies, we stopped and counted that we collected 30 unique quotes.

*

*What is the expected value of N?

*What is the most probable value of N?

 A: These questions are not well-defined as posed. We need to know the prior probability distribution for $N$ in order to derive the posterior probability distribution using the information that you collected $30$ unique quotes.
Let’s assume an improper uniform prior for $N$ (improper because it can’t be normalized, as there’s no uniform distribution on $\mathbb N$).
The probability that you get $k=30$ unique quotes out of $m=100$ from opening $N$ cookies is
$$
\def\stir#1#2{\left\{#1\atop#2\right\}}
\binom mk\left(\frac km\right)^N\left(\frac{k!}{k^N}\right)\stir Nk\;,
$$
since you can choose the $k$ unique quotes in $\binom mk$ ways, the probability to get only quotes among those $30$ in $N$ draws is $\left(\frac km\right)^N$, and the probability that all $30$ are covered after $N$ draws is $\left(\frac{k!}{k^N}\right)\stir Nk$. The factors $k^N$ cancel, and we can drop all factors that don’t depend on $N$, yielding
$$
m^{-N}\stir Nk\;.
$$
For a uniform prior, we can normalize this likelihood to obtain the posterior distribution for $N$. The generating function for the upper index of the Stirling numbers of the second kind is (see Wikipedia)
$$
\sum_{n=k}^\infty\stir nkx^{n-k}=\prod_{r=1}^k\frac1{1-rx}\;,
$$
so we have
$$
\sum_{N=k}^\infty m^{-N}\stir Nk=m^{-k}\prod_{r=1}^k\frac1{1-\frac rm}
$$
and
\begin{eqnarray}
\sum_{N=k}^\infty Nm^{-N}\stir Nk
&=&
-m\frac{\mathrm d}{\mathrm dm}\sum_{N=k}^\infty m^{-N}\stir Nk
\\
&=&
-m\frac{\mathrm d}{\mathrm dm}\left(m^{-k}\prod_{r=1}^k\frac1{1-\frac rm}\right)
\\
&=&
\left(m^{-k}\prod_{r=1}^k\frac1{1-\frac rm}\right)\left(k+\sum_{r=1}^k\frac r{m-r}\right)
\\
&=&
\left(m^{-k}\prod_{r=1}^k\frac1{1-\frac rm}\right)\left(\sum_{r=1}^k\frac m{m-r}\right)\;,
\end{eqnarray}
so the expected value of $N$ (assuming a uniform prior) is
$$
m\sum_{r=1}^k\frac1{m-r}\;.
$$
In your case, this is about $35.88$ (Wolfram|Alpha computation). Note that this differs from the approximation you got from the expected number of draws required to obtain $k$ unique quotes by $m\left(\frac1{m-k}-\frac1m\right)=\frac k{m-k}$. See Why is the expected number of draws given $k$ coupons so similar to the expected number of draws required to obtain $k$ coupons? for a more direct derivation of this result without the detour via generating functions.
From this plot (Wolfram|Alpha computation), you can see that the most likely value of $N$ (assuming a uniform prior) is $35$:

