Probability question with 4 dice. 
1) Roll for 4 die. What is the probability of throwing two distinct pairs of dice.

I think I solved this by considering each roll as we have independent events so let $A$ be the event of 2 pairs of distinct pairs
$$
\mathbb{P}(A) = \mathbb{P}(A_1 \cap A_2 \cap A_3 \cap A_4)= {3 \choose 2}\cdot\mathbb{P}(A_1)\cdot\mathbb{P}(A_2)\cdot\mathbb{P}(A_3)\cdot\mathbb{P}(A_4)
$$ 
where:


*

*$A_1$ is the first roll, so it can be any one of the 6 faces.

*$A_2$ is the second roll, and has probability $1/6$.

*$A_3$ can be any number but that of $A_1$ and $A_2$ so has probability $5/6$.

*$A_4$, like the second roll, can only be one of 6 numbers, so has probability $1/6$.


Since there are 3 choose 2 ways of arranging our rolls, the total probability is
$$
\mathbb{P}(A) = 3\cdot 1 \cdot \frac 1 6 \cdot \frac 5 6 \cdot \frac 1 6 = \frac{5}{72}.
$$
Now more interestingly:

Generalise this game to multiple rounds. Each round we can remove all the dice, a pair, or no dice and then re-roll the dice we have removed from the game. The game ends when we have 2 pairs of distinct pairs of die. What is the probability that after say, $3$ rounds, we have 2 distinct pairs of die.

Now I assume the optimum strategy is if we don't have two distinct pairs, we remove either 1 pair of unpaired dice if the other two are paired, or we remove all dice. However I'm not sure how to go from here.
 A: For the second part of the question, we have $4$ different states: nothing (the initial state), one pair, one pair and a distinct singleton, and two distinct pairs (the absorbing winning state). Let’s number them $0$ through $3$ in that order. Then the transition matrix is
$$
6^{-4}\pmatrix{
6\cdot5\cdot4\cdot3&0&0&0\\
6\cdot1\cdot1\cdot1&6^2\cdot1\cdot1&0&0\\
6\cdot5\cdot4\cdot\binom42+6\cdot5\cdot4&6^2\cdot6\cdot5&6^3\cdot5&0\\
6\cdot5\cdot\binom32&6^2\cdot5&6^3\cdot1&6^4}
=
6^{-3}\pmatrix{
60&0&0&0\\
1&6&0&0\\
140&180&180&0\\
15&30&36&216
}\;.
$$
We can let Wolfram|Alpha compute the third power of this matrix:
$$
6^{-7}\pmatrix{
6000&0&0&0\\
111&6&0&0\\
183230&167580&162000&0\\
90595&112350&117936&279936
}\;.
$$
The probability that the absorbing winning state is reached in three rounds from the initial state is the bottom left entry,
$$
\frac{90595}{279936}\approx0.3236\;.
$$
Roughly, you have a $1$ in $3$ chance to get two distinct pairs, a $2$ in $3$ chance to get one pair and a distinct singleton, and a negligible chance to get nothing or only one pair.
