In Maxi Yahtzee (Yahtzee with 6 dices): How to calculate the probability of getting two trios in one round?. Idem for three duos.
Rules:
The game consists of a number of rounds. In each round, a player gets three rolls of the dice, although they can choose to end their turn after one or two rolls. After the first roll the player can save any dice they want and re-roll the other dice. This procedure is repeated after the second roll. The player has complete choice as to which dice to roll. They can re-roll a die for the third roll that was not rolled on the second roll.
Note: 333444 is a double trio. But 333333 it is not.
334455 is a triple duo. But 333355 and 333333 they are not.
Best Answer
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
$$\frac{221494355}{1088391168} \approx 0.2035$$