From a 32-card Hungarian deck, we draw 6 cards. What is the probability that all four suits appear in these 6 cards? What is this probability in the case of a 52-card French deck?
The composition of the cards is favorable if the distribution by colors is I.) 3, 1, 1, 1 or II.) 2, 2, 1, 1. (In any other distribution, at least one color is missing.
a) In the case of a 32-card Hungarian deck:

In the first case, the number of favorable cases is clearly

$$
k_{\mathrm{I}}=\binom{4}{1}\binom{8}{3}\binom{8}{1}^{3}=4 \cdot \frac{8 \cdot 7 \cdot 6}{1 \cdot 2 \cdot 3} \cdot 8^{3}=4 \cdot 7 \cdot 8^{4}
$$

In the second case,

$$
k_{\mathrm{II}}=\binom{4}{2}\binom{8}{2}^{2}\binom{8}{1}^{2}=6\left(\frac{8 \cdot 7}{1 \cdot 2}\right)^{2} 8^{2}=3 \cdot 4 \cdot 7^{2} \cdot 8^{3}
$$

The number of all possible cases is:

$$
l=\binom{32}{6}=\frac{32 \cdot 31 \cdot 30 \cdot 29 \cdot 28 \cdot 27}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}=4 \cdot 31 \cdot 29 \cdot 28 \cdot 9
$$

and thus the sought probability is

$$
v_{a}=\frac{k_{\mathrm{I}}+k_{\mathrm{II}}}{l}=\frac{4 \cdot 7 \cdot 8^{3}(8+3 \cdot 7)}{4 \cdot 31 \cdot 29 \cdot 28 \cdot 9}=\frac{2 \cdot 8^{2}}{9 \cdot 31}=\frac{128}{279} \approx 0.459
$$

b) In the case of a 52-card French deck:

$$
\begin{aligned}
k_{\mathrm{I}} & =\binom{4}{1}\binom{13}{3}\binom{13}{1}^{3}=4 \cdot \frac{13 \cdot 12 \cdot 11}{1 \cdot 2 \cdot 3} 13^{3}=8 \cdot 11 \cdot 13^{4} \\
k_{\mathrm{II}} & =\binom{4}{2}\binom{13}{2}^{2}\binom{13}{1}^{2}=6\left(\frac{13 \cdot 12}{1 \cdot 2}\right)^{2} 13^{2}=3 \cdot 6 \cdot 12 \cdot 13^{4}=8 \cdot 27 \cdot 13^{4} \\
l & =\binom{52}{6}=\frac{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48 \cdot 47}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}=52 \cdot 17 \cdot 10 \cdot 49 \cdot 47
\end{aligned}
$$

and thus

$$
v_{b}=\frac{8 \cdot 13^{4}(11+27)}{52 \cdot 17 \cdot 10 \cdot 49 \cdot 47}=\frac{2 \cdot 13^{3} \cdot 19}{17 \cdot 5 \cdot 47 \cdot 49}=\frac{83486}{195755} \approx 0.426
$$

András Bauer (Bp. II., Rákóczi g. IV. o. t.)