Four people sit at the four sides of a table and the 32-card Hungarian deck is dealt out equally among them. If one of the selected players does not get an ace, what is the probability that at least one other player has exactly two (no more, no less) aces?
I. solution: The selected player's hand does not contain an ace, so the other 3 players share 24 cards, among which are four aces and 20 non-aces. Let's calculate the opposite probability, that is, the probability that none of the three players get two aces. The latter case can only occur with the distribution 4, 0, 0 or $3,1,0$.

When one player gets four aces, the number of favorable cases is the same as the number of 4th class combinations that can be formed from the 20 non-ace cards, which is $\binom{20}{4}$. Considering that any of the three players can get the four aces, the number of favorable cases is $\binom{3}{1}\binom{20}{4}$. The number of possible cases is $\binom{24}{8}$, and thus the probability that one of the three players gets four aces is

$$
\begin{gathered}
v_{4}=\frac{\binom{3}{1}\binom{20}{4}}{\binom{24}{8}}=\frac{3 \cdot 20 \cdot 19 \cdot 18 \cdot 17 \cdot 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 24 \cdot 23 \cdot 22 \cdot 21 \cdot 20 \cdot 19 \cdot 18 \cdot 17}= \\
\frac{3 \cdot 5}{23 \cdot 11 \cdot 3}=\frac{15}{759}
\end{gathered}
$$

Similarly, the probability that one of the 3 players gets 3 aces is

$$
\begin{gathered}
v_{3}=\frac{\binom{3}{1}\binom{4}{3}\binom{20}{5}}{\binom{24}{8}}=\frac{3 \cdot 4 \cdot 20 \cdot 19 \cdot 18 \cdot 17 \cdot 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 24 \cdot 23 \cdot 22 \cdot 21 \cdot 20 \cdot 19 \cdot 18 \cdot 17}= \\
=\frac{2 \cdot 16 \cdot 6}{23 \cdot 11 \cdot 3}=\frac{192}{759}
\end{gathered}
$$

Thus, the probability sought in the problem is

$$
1-\left(v_{4}+v_{3}\right)=1-\frac{207}{759}=\frac{552}{759}=\frac{8}{11} \approx 0.727
$$

Péter Kiss (Vak Bottyán g. III. o. t.)

II. solution: The desired probability can also be calculated directly.

The favorable distributions are: $2,2,0$ and $2,1,1$.

The probability of the first case is

$$
v_{2,2}=\binom{3}{1} \frac{\binom{4}{2}\binom{20}{6}}{\binom{24}{8}} \cdot \frac{\binom{14}{6}}{\binom{16}{8}}=3 \cdot \frac{240}{759} \cdot \frac{7}{30}=\frac{168}{759}
$$

The probability of the $2, 1, 1$ distribution is (considering the previous partial result)

$$
v_{2,1}=3 \cdot \frac{240}{759}\binom{2}{1} \frac{\binom{14}{7}}{\binom{16}{8}}=3 \cdot \frac{240}{759} \cdot \frac{8}{15}=\frac{384}{759}
$$

Thus, the desired probability is

$$
\begin{aligned}
v_{2,2}+v_{2,1}=\frac{168}{759}+\frac{384}{759}= & \frac{552}{759}=\frac{8}{11} \\
& \text { László Kovács (Debrecen Református g. IV. o. t.) }
\end{aligned}
$$

III. solution: It is known that

$$
v_{A / B}=\frac{v_{A B}}{v_{B}}
$$

where $B$ represents that one player has no aces,

$A$ represents that one player has 2 aces,

$A B$ represents the simultaneous occurrence of the two events.

In case $B$, the number of favorable cases for a specific player is the number of ways to choose 8 cards from the 28 non-ace cards: $\binom{28}{8}$, so the total number of favorable cases is $\binom{4}{1}\binom{28}{8}$.

In case $A B$, note that the fact that one player has exactly two aces already implies that at least one player has no aces. Therefore, every case where one player has exactly two aces is favorable. Thus, the number of favorable cases in case $A B$ is $\binom{4}{1}\binom{4}{2}\binom{28}{6}$.

Thus,

$$
v_{A B}=\frac{\left(\begin{array}{l}
4 \\
1 \\
1
\end{array}\right)\binom{4}{2}\binom{28}{6}}{\binom{32}{8}}, \quad v_{B}=\frac{\binom{4}{1}\binom{28}{8}}{\binom{32}{8}}
$$

and thus

$$
v_{A / B}=\frac{\binom{4}{2}\binom{28}{6}}{\binom{28}{8}}=\frac{8}{11}
$$

István Gaál (Csorna, Latinka S. g. IV. o. t.)