Calculating probability of getting 3 pairs in 13 cards(similar and related to another question I asked before) Assuming there are 52 cards and dealt with 4 different players. Each of them will receive 13 cards. What are the possibilities of a player receive exactly 3 pairs?
*four-of-a-kind is considered 2 pairs.
This means we have  3 different pairs and 2 same pairs and one different pair.
For 3 different pairs,
${{13}\choose{2}}$ is for two ranks showing on the two pairs.
$ {{10}\choose{7}}$ is with 3 ranks being chosen and 10 ranks left, we choose from there.
$ {{4}\choose{2}}$ is the combination for the pair in the rank.
$ {{4}\choose{1}}$ is the suit for the non-matching cards.
For 2 same pairs and one different pair,
${{13}\choose{2}}$ is for two ranks showing on the two pairs.
$ {{10}\choose{7}}$ is with 2 ranks being chosen and 11 ranks left, we choose from there.
$ {{4}\choose{4}}$ is the combination for the 2 same pairs in the rank.
$ {{4}\choose{2}}$ is the combination for the pair in the rank.
$ {{4}\choose{1}}$ is the suit for the non-matching cards.
$$
P=\frac{\binom{13}{2}\cdot\binom{10}{7}\cdot\binom{4}{2}\cdot\binom{4}{2}\cdot\binom{4}{2}\cdot {{4}\choose{1}}^7}{\binom{52}{13}}+\frac{\binom{13}{2}\cdot\binom{11}{7}\cdot\binom{4}{4}\cdot\binom{4}{2}\cdot\binom{4}{1}^7}{\binom{52}{13}}=1.3707 \times{10}^{-5}
$$
Does this way of calculating probability include both situations or it doesn't?
 A: I agree with your count with the following remarks.
For 3 different pairs instead of “${{13}\choose{2}}$ is for two ranks showing on the two pairs”
should be “${{13}\choose{3}}$ is for three ranks showing on the three pairs”.
For 2 same pairs and one different pair instead of “${{13}\choose{2}}$ is for two ranks showing on the two pairs” should be “$13$ is for the rank showing on the doubled pair and $12$ is for the rank showing on the remaining pair” and instead of “$ {{10}\choose{7}}$” should be $ {{11}\choose{7}}$.
A: This way of calculating is correct in the sense that it accounts for all possibilities and it does not double count some possibilities, but...
Apart from the typing mistakes which Alex already pointed out, there is also a conceptual error in the proposed solution.
When you have 3 pairs in different ranks, they can be treated as equal, so you have $\binom{13}{3}\cdot\binom{4}{2}^3$ combinations to take these pairs.
When you have 2 pairs in the same rank and 1 in another, the third pair can be distinguished from the first two, so there are more choices than $\binom{13}{2}$. It is easier to count these combinations as $13\cdot 12 \cdot \binom{4}{4}\cdot\binom{4}{2}$. This would give the following probability:
$$
P=\frac{\binom{13}{3}\cdot\binom{10}{7}\cdot\binom{4}{2}^3\cdot {{4}\choose{1}}^7}{\binom{52}{13}}+\frac{13\cdot12\cdot\binom{11}{7}\cdot\binom{4}{4}\cdot\binom{4}{2}\cdot\binom{4}{1}^7}{\binom{52}{13}} \approx 0.199
$$
Note also that this is the probability that one fixed player receives 3 pairs. The probability that at least one player gets 3 pairs, would be significantly harder to compute, since they cannot be considered independent. (This is how I interpreted the question at first.)
