In a 52-card French deck, there are four suits, each with 13 cards of different values. Which of the following is more likely:
a) 10 cards of a specific suit out of 13 dealt cards,

b) the 13 dealt cards representing 13 different values.
The number of possible cases in both scenarios is $\binom{52}{13}$. In case a), the 10 specified cards of the same suit can be chosen from the 13 cards of the same suit in $\binom{13}{10}$ ways. The other 3 cards can be chosen from the 39 cards of different suits in $\binom{39}{3}$ ways. Therefore, the number of favorable cases is $\binom{13}{10}\binom{39}{3}$.

In case b), each value can be chosen in 4 ways, so 13 cards of different values can be chosen in $4^{13}$ ways. (Each selection is essentially a 13-class repeated permutation of the 4 suits.)

Let's compare the probabilities of a) and b):

$$
\frac{v_{a}}{v_{b}}=\frac{\binom{13}{3}\binom{39}{3}}{4^{13}}=\frac{\frac{13 \cdot 12 \cdot 11}{1 \cdot 2 \cdot 3} \cdot \frac{39 \cdot 38 \cdot 37}{1 \cdot 2 \cdot 3}}{2^{26}}=\frac{13^{2} \cdot 11 \cdot 38 \cdot 37}{2^{26}}
$$

In the numerator, replace each factor with a power of 2 that is already greater than the respective number:

$$
\frac{v_{a}}{v_{b}}<\frac{\left(2^{4}\right)^{2} \cdot 2^{4} \cdot 2^{6} \cdot 2^{6}}{2^{26}}=\frac{2^{24}}{2^{26}}<1
$$

Thus, $v_{a}<v_{b}$, and therefore case b) is more probable.

Zsombok Zoltán (Bp., IV., Konyves Kalman g. II. o. t.)