In the case of two triples, the Transition matrix $P_{33}$ is given by
$$
P_{33} =
\left(
\begin{array}{ccccccc}
\frac{5}{324} & \frac{25}{108} & \frac{1531}{7776} & \frac{125}{324} & \frac{425}{2592} & \frac{25}{3888}\\
0 & \frac{5}{54} & \frac{127}{432} & \frac{35}{108} & \frac{355}{1296} & \frac{5}{324}\\
0 & 0 & \frac{17}{27} & 0 & \frac{25}{72} & \frac{5}{216}\\
0 & 0 & 0 & \frac{4}{9} & \frac{1}{2} & \frac{1}{18}\\
0 & 0 & 0 & 0 & \frac{5}{6} & \frac{1}{6}\\
0 & 0 & 0 & 0 & 0 & 1
\end{array}
\right)$$
The $p_{i,j}$ element represents the probability of moving from state i to j, according to the following table
\begin{array}{ccccccc}
State\\
1: & 0 \:pair, 0 \:triple\\
2: & 1 \:pair, 0 \:triple\\
3: & 0 \:pair, 1 \:triple\\
4: & 2 \:pairs\\
5: & 1 \:pair, 1 \:triple\\
6: & 2 \:triples
\end{array}
Then the probability of getting two triples in 3 or less rolls is given by the sixth element of
$$\left(
\begin{array}{ccccccc}
1 & 0 & 0 & 0 & 0 & 0
\end{array}
\right)\cdot P_{33}^{3}$$
that is to say
$$\frac{260649145}{1632586752} \approx 0.1597$$
In the case of three pairs, the Transition matrix $P_{222}$ is given by
$$
P_{222} =
\left(
\begin{array}{ccccccc}
\frac{5}{324} & \frac{3331}{7776} & \frac{4025}{7776} & \frac{25}{648}\\
0 & \frac{167}{432} & \frac{245}{432} & \frac{5}{108}\\
0 & 0 & \frac{8}{9} & \frac{1}{9}\\
0 & 0 & 0 & 1
\end{array}
\right)
$$
The $p_{i,j}$ element represents the probability of moving from state i to j, according to the following table
$$
\begin{array}{ccccccc}
State\\
1: & 0 \:pair\\
2: & 1 \:pair\\
3: & 2 \:pairs\\
4: & 3 \:pairs
\end{array}
$$
Then the probability of getting three pairs in 3 or less rolls is given by the fourth element of
$$
\left(
\begin{array}{ccccccc}
1 & 0 & 0 & 0
\end{array}
\right)\cdot P_{222}^{3}
$$
that is to say
$$\frac{221494355}{1088391168} \approx 0.2035$$
444444
and444433
count as three pairs? $\endgroup$