I have a problem solving the following:
Out of a $52$-card deck, I'm missing all my kings and one card of number $2$. Now if I pull five cards randomly, what is the chance to get $4$ of a kind?
Because I am missing $4$ kings and one card of number $2$, it remains only $11$ out of $13$ values to get $4$ of a kind from. When we have chosen our value, we can choose the last ($5^\text{th}$) card in $11 \times 43 = 473$ different ways (according to the multiplication principle). $11$ is the values that remain and $43$ is the cards that remain after I have pulled $4$ cards. $47$ cards from the start (because I am missing $5$ cards). The total of different poker hands is $\displaystyle \binom{47}{5}$ and the probability to get $4$ of a kind is therefore
$$(11 \times 43) \div \dfrac{47 \times 46 \times 45 \times 44 \times 43}{1 \times 2 \times 3 \times 4 \times 5} = \frac{1 \times 2 \times 3 \times 4 \times 5 \times 11}{47 \times 46 \times 45 \times 44} = \frac{1 \times 2 \times 3 \times 4 \times 5 \times 11}{47 \times (2 \times 23) \times (5 \times 3 \times 3) \times (4 \times 11)} = \frac{1}{47 \times 23 \times 3} = 0,0003 = 0,03 \%$$
Is this a correct calculation?
Also if I would drop one more card, which card should I drop to maximize my chances so that $5$ randomly cards give $4$ of a kind?
Directly, I would say that I could drop one more of the number $2$ because from the begging I only had $3$ cards of the number $2$ and therefore I could not get any $4$ of a kind from number $2$ anyways. Is it correct or should I think otherwise?
I am thankful for any tips and advice.
Yes, your approach and results are correct.