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.
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.