Four people sit at a table, one on each side, and they deal the 32-card Hungarian deck equally among themselves. If one of the selected players does not get an ace, what is the probability that the player sitting opposite them also does not have an ace in their 8 cards?
This is about conditional probability. The question is, what is the probability that someone does not get an ace (event $A$) in 8 cards, given that another player also did not get an ace (event $B$) in their 8 cards.

The number of possible cases is the number of 8-combinations that can be formed from the remaining 24 cards, which is $\binom{24}{8}$. Favorable cases are all 8-combinations that can be formed from the 20 cards remaining after excluding the 4 aces, so their number is $\binom{20}{8}$, and thus the sought probability is

$$
V_{A / B}=\frac{\binom{20}{8}}{\binom{24}{8}}=\frac{20 \cdot 19 \ldots 13}{24 \cdot 23 \ldots 17}=\frac{16 \cdot 15 \cdot 14 \cdot 13}{24 \cdot 23 \cdot 22 \cdot 21}=\frac{2 \cdot 5 \cdot 13}{3 \cdot 23 \cdot 11}=\frac{130}{759} \approx 0.171
$$

Deseő Zoltán (Bp., X., I. László g. III. o. t.)

Notes: 1. The unconditional, absolute probability

$$
V_{A}=\frac{\binom{28}{8}}{\binom{32}{8}}=\frac{24 \cdot 23 \cdot 22 \cdot 21}{32 \cdot 31 \cdot 30 \cdot 29}=\frac{5313}{16980} \approx 0.313
$$

2. When the conditional probability can be easily calculated directly, there is no need to use the formula $V_{A / B}=\frac{V_{A B}}{V_{B}}$.