Three players (András, Béla, and Csaba) are playing with a deck of 32 Hungarian cards.
10 cards are dealt to each player, and the remaining 2 cards go into a separate pile (the talon).
There are eight ranks (7, 8, 9, 10, J, Q, K, A) in the Hungarian deck, with 4 cards of each rank (one from each suit).
We need to find the probability that András does not receive all four cards of any rank (i.e., he does not have four of a kind in any rank).

To solve this problem, we need to compute the probability that András does not receive all four cards of any rank (i.e., he does not have four of a kind in any rank) from a deck of Hungarian cards, which has 32 cards in total.

### Step 1: Total Possible Hands for András
There are 32 cards in the deck, and each player (including András) is dealt 10 cards. The total number of ways to deal 10 cards to András is the number of ways to choose 10 cards from 32 cards. This can be calculated using the binomial coefficient:

\[
\binom{32}{10} = \frac{32!}{10!(32 - 10)!} = \frac{32!}{10!22!}
\]

### Step 2: Number of "Unwanted" Hands for András
We now want to compute the number of hands in which András receives **all four cards** of at least one rank. To do this, we'll first compute the number of hands in which András receives all four cards of exactly one specific rank.

For each rank, there are 4 cards, and András will receive all of them. The remaining 6 cards must be chosen from the other 28 cards (since we've removed the 4 cards of the chosen rank). The number of ways to select 6 cards from the remaining 28 cards is:

\[
\binom{28}{6}
\]

Thus, the number of ways András can receive all four cards of a particular rank is:

\[
\binom{28}{6}
\]

Since there are 8 ranks in the deck, the total number of ways András can receive four of a kind in any rank (with no further restrictions) is:

\[
8 \times \binom{28}{6}
\]

### Step 3: Subtract the "Unwanted" Hands
The number of valid hands, where András does not receive four of a kind in any rank, is the total number of hands minus the number of unwanted hands (where he receives four of a kind in at least one rank):

\[
\text{Valid hands} = \binom{32}{10} - 8 \times \binom{28}{6}
\]

### Step 4: Compute the Probability
The probability that András does **not** receive four of a kind in any rank is the ratio of valid hands to the total number of hands:

\[
P = \frac{\binom{32}{10} - 8 \times \binom{28}{6}}{\binom{32}{10}}
\]

Now we can compute the values:

- \(\binom{32}{10}\)
- \(\binom{28}{6}\)

Let me compute those.The probability that András does **not** receive all four cards of any rank is approximately \( 0.953 \), or 95.3%.

Would you like more details on any part of the process?

Here are 5 related questions you might find interesting:
1. What is the probability that Béla receives all four cards of one specific rank?
2. How does the probability change if there were only 30 cards in the deck?
3. What is the probability that both András and Béla do not receive all four cards of any rank?
4. How would the probability be affected if each player received 9 cards instead of 10?
5. How can we calculate the probability that András gets exactly one "four of a kind"?

**Tip:** The use of binomial coefficients is a common technique in combinatorics, especially for problems involving selection without replacement!

Three players (András, Béla, and Csaba) are playing with a deck of 32 Hungarian cards. 10 cards are dealt to each player, and the remaining 2 cards go into a separate pile (the talon). There are eight ranks (7, 8, 9, 10, J, Q, K, A) in the Hungarian deck, with 4 cards of each rank (one from each suit). We need to find the probability that András does not receive all four cards of any rank (i.e., he does not have four of a kind in any rank).

Explore the detailed probability calculation of András not receiving four of a kind from a deck of 32 Hungarian cards. This problem involves combinatorial concepts and the use of binomial coefficients to compute probabilities, making it ideal for advanced students in combinatorics and probability theory.

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