We draw three cards from a Hungarian deck of cards.

a) If all three cards are of different suits, what is the probability that at least one of the three cards is an ace?

b) What is the probability of the latter, without any conditions?

c) What is the probability that we draw three cards of different suits, among which at least one is an ace?
# I. Solution:

a) Calculate the probability $\left(v_{A / B}\right)$ that there is no ace among the 3 cards, given that the 3 cards are of different suits.

So

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

where $v_{A B}$ is the probability that all 3 cards are of different suits and none of them is an ace, and $v_{B}$ is the probability that all 3 cards are of different suits.

In the case of $v_{A B}$, the favorable outcomes are any 3rd class combination of the 4 suits and any repetitive variation of the 7 values excluding the ace for each combination. Therefore, the number of favorable outcomes is $\binom{4}{3} 7^{3}$. The total number of possible outcomes is $\binom{32}{3}$, so

$$
v_{A B}=\frac{\binom{4}{3} 7^{3}}{\binom{32}{3}}
$$

Similarly,

$$
v_{B}=\frac{\binom{4}{3} 8^{3}}{\binom{32}{3}}
$$

and thus

$$
v_{A / B}=\frac{v_{A B}}{v_{B}}=\frac{7^{3}}{8^{3}}=\frac{343}{512}
$$

Therefore, the sought probability is

$$
V_{\alpha}=1-v_{A / B}=1-\frac{343}{512}=\frac{169}{512} \approx 0.330
$$

b) If we are looking for the probability $v^{\prime}$ that there is no ace among the 3 cards, then the favorable outcomes are any 3rd class combination of the 28 cards excluding the 4 aces, i.e.,

$$
v^{\prime}=\frac{\binom{28}{3}}{\binom{32}{3}}=\frac{28 \cdot 27 \cdot 26}{32 \cdot 31 \cdot 30}=\frac{7 \cdot 9 \cdot 13}{4 \cdot 31 \cdot 10}=\frac{819}{1240}
$$

and thus the sought probability is

$$
v_{b}=1-v^{\prime}=1-\frac{819}{1240}=\frac{421}{1240} \approx 0.340
$$

c) If we consider the sought probability $v_{c}$ as $v_{A B}$, then

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

where $v_{A / B}$, based on part a), is $\frac{169}{512}$ and $v_{B}$ is the probability of drawing 3 cards of different suits, which, as we saw,

$$
\frac{\binom{4}{3} 8^{3}}{\binom{32}{3}}=\frac{4 \cdot 8^{3} \cdot 6}{32 \cdot 31 \cdot 30}=\frac{8^{2}}{31 \cdot 5}=\frac{64}{155}
$$

and thus the sought probability is

$$
v_{c}=v_{A B}=\frac{169}{512} \cdot \frac{64}{155}=\frac{169}{1240} \approx 0.136
$$

Note: $V_{B}$ can also be calculated this way: The probability that the second card drawn is of a different suit than the first is $\frac{24}{31}$, and the probability that the third card drawn is of a different suit than the first two is $\frac{16}{30}$. Therefore, $v_{B}=\frac{24}{31} \cdot \frac{16}{30}=\frac{64}{155}$.

II. Solution:

a) If we always draw a card of a different suit, then the probability of not drawing an ace in each draw is $\frac{7}{8}$, so by the multiplication rule, the probability of not drawing an ace in 3 draws, given that each draw is of a different suit, is $\left(\frac{7}{8}\right)^{3}$. Therefore, the sought probability is

$$
v_{\alpha}=1-\left(\frac{7}{8}\right)^{3}=1-\frac{343}{512}=\frac{169}{512}
$$

b) The probability of not drawing an ace on the first draw is $\frac{28}{32}$, on the second draw is $\frac{27}{31}$, and on the third draw is $\frac{26}{30}$, so the probability of not drawing an ace in 3 draws is

$$
\frac{28}{32} \cdot \frac{27}{31} \cdot \frac{26}{30}=\frac{819}{1240}
$$

Therefore, the sought probability is

$$
v_{b}=1-\frac{819}{1240}=\frac{421}{1240}
$$

c) The number of possible cases for 3 different suits, as we saw, is $\binom{4}{3} 8^{3}$. Among these, the number of cases without an ace is $\binom{4}{3} 7^{3}$, so the number of cases with at least one ace is $\binom{4}{3} 8^{3}-\binom{4}{3} 7^{3}$, and thus the sought probability is

$$
v_{c}=\frac{4\left(8^{3}-7^{3}\right)}{\binom{32}{3}}=\frac{4\left(8^{3}-7^{3}\right) 6}{32 \cdot 31 \cdot 30}=\frac{8^{3}-7^{3}}{8 \cdot 31 \cdot 5}=\frac{169}{1240}
$$

Géza Biczó (Bp., II., Rákóczi g. II. o. t.) Imre Csiszár (Bp., I., Petőfi g. I. o. t.)