To solve this problem, we need to calculate the probability that, given all five cards drawn are number cards and there are at least two cards of the same number, only two different numbers are obtained, with at most three cards of the same number.

Step-by-Step Solution:

1. Identify Total Number Cards and Constraints:

   • There are 10 number cards (Ace through 10), each with 4 suits, so 40 number cards in total.
   • Five cards are drawn, all of which are number cards.

2. Define Desired Event:

   We want to find the probability that:

   • Only two different numbers are drawn.
   • At most three cards of the same number are drawn.

3. Count Favorable Outcomes:

   Let $a$ and $b$ represent the two different numbers drawn. We need to count the combinations where the distribution of the five cards is:

   • Either (3,2) or (2,3) in terms of the number of cards of each type.

   Case (3,2):

   • Select 2 out of 10 numbers: $\binom{10}{2}$
   • Choose 3 cards from one number (4 possible suits): $\binom{4}{3} = 4$
   • Choose 2 cards from the other number (4 possible suits): $\binom{4}{2} = 6$

   The number of favorable ways in this case is: $\binom{10}{2} \times \binom{4}{3} \times \binom{4}{2} = 45 \times 4 \times 6 = 1080$

   Case (2,3):

   • This is essentially the same as (3,2), just swapped.

   So, the number of favorable ways for the entire event is: $1080 + 1080 = 2160$

4. Total Number of Outcomes:

   • Since we know all cards drawn are number cards, we have: $\binom{40}{5}$

5. Probability Calculation:

   The probability is the ratio of favorable outcomes to the total outcomes: $P = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Outcomes}} = \frac{2160}{\binom{40}{5}}$

6. Calculating $\binom{40}{5}$:

   $\binom{40}{5} = \frac{40!}{5!(40-5)!} = \frac{40!}{5! \cdot 35!}$

   Let's compute this value: $\binom{40}{5} = \frac{40 \times 39 \times 38 \times 37 \times 36}{5 \times 4 \times 3 \times 2 \times 1} = 658008$

7. Final Probability:

   $P = \frac{2160}{658008} \approx 0.00328$

So, the probability that only two different numbers are obtained from the cards drawn and at most three cards of the same number are drawn is approximately $0.00328$.

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Do you want more details or have any questions?

Here are some related questions:

1. How would the probability change if we allowed up to four cards of the same number?
2. What is the probability of drawing exactly three different numbers?
3. How many total ways can five cards be drawn such that all are number cards?
4. What is the probability of drawing five cards all of different numbers?
5. How does the probability change if the deck has jokers included?
6. What is the probability of drawing a full house (three of one number and two of another) from number cards?
7. How does the probability change if only four cards are drawn?
8. What is the probability of drawing two pairs and a different card?

Tip: Always simplify the problem by breaking it into smaller steps, especially when dealing with probabilities and combinatorial counts.