I found an interesting question in a textbook for school kids:
Estimate the minimum number of eligible voters that one must interview in order to be able to predict with a probability $\ge90\%$ the election result for a particular party with an error of $\le1\%$.
I think this must be solved applying multinomial or multivariate hypergeometric distributions with unknown number of parties.
The question is translated from a German textbook and the original quote is as follows:
Man schätze die Anzahl der Wahlberechtigten, die man mindestens befragen muß, um mit einer Wahrscheinlichkeit von 90% oder mehr das Wahlergebnis für eine bestimmte Partei mit einem Fehler von höchstens 1% vorhersagen zu können.
Franz Heigl, Jürgen Feuerpfeil: Stochastik. ein Lehr- und Arbeitsbuch, München 1987, p.179
No fancy distributions are required for a problem like this; it is always assumed that the sample size is large enough that successive draws from the population can be regarded as approximately independent. In particular, we don't need to reach for a multinomial distribution, because by focusing on one party we have translated the problem to a binomial consideration (a respondent either does, or does not, prefer that party).
At its core, this is about deriving a $90\%$ confidence interval for a population proportion $p$. By a common statistical result, the confidence interval has the form $$\hat p \pm z_{\alpha / 2} \sqrt{ p (1-p) / n}$$ where $\hat p$ is the sample estimate of the proportion, $z_{\alpha/2}$ is a critical value from the normal distribution (in this case, $1.645$), $p$ is the true population proportion, and $n$ is the sample size. The margin of error is the term on the right side of the $\pm$ sign, so our task is to make $$1.645 \sqrt{p (1 - p)} / \sqrt n < 0.01. $$ Note that the worst case for the expression $\sqrt{p (1 - p)}$ occurs when $ p = 1/2$, so we need $$\sqrt n > \frac{1.645 \cdot \sqrt{(1/2)^2}}{0.01}$$ and solving for $n$ gives an answer of $\fbox{6766}$ people (rounding up to guarantee we see the desired confidence).