Probability of $3+3$ cards, out of $6$ cards drawn from a solitaire 
A solitaire consists of $52$ cards. We take out $6$ out of them (wihout repetition). Find the probability there are $3+3$ cards of the same type (for example, $3$ "1" and $3$ "5").

Attempt. 
First approach. There are $\binom{13}{2}$ ways to choose $2$ out of the $13$ types and by the multiplication law of probability, the desired probability is $$\binom{13}{2}\frac{4}{52}\,\frac{3}{51}\,\frac{2}{50}\,
\frac{4}{49}\,\frac{3}{48}\,\frac{2}{47}.$$
Second approach. There are $\binom{13}{2}$ ways to choose $2$ out of the $13$ types and the desired probability is $$\binom{13}{2}\frac{\binom{4}{3}\binom{4}{3}\binom{4}{0}\ldots\binom{4}{0}}{\binom{52}{6}}.$$
These numbers don't coincide, so I guess (at least) one of them is not correct.
Thanks in advance for the help.

There are $13$ possible numbers/people. We need to pick $3$, so the probability is  $\binom{4}{3} \cdot \binom{13}{2} \cdot \binom{4}{3}$. The total probability of choosing 6 cards is  $\binom{13}{2}$. The total probability is $\frac{4 \cdot 78 \cdot 4}{20358520}$
, which equals $\frac{1248}{20358520} \Rightarrow \approx 6.13011162*10^{-5}$
