We have a drawing with replacement of $10$ balls out of an urn containing $6$ red balls, numbered from $1$ to $6$, and 30 white balls.
The number $R$ of red balls drawn is binomially distributed; i.e., the probability $p_r$ that we draw exactly $r$ red balls is given by
$$p_r={10\choose r}\left({1\over6}\right)^r\left({5\over 6}\right)^{10-r}\qquad(0\leq r\leq 10)\ .$$
Assuming that we draw exactly $r$ red balls, the numbers on these $r$ balls define a map $f:\ [r]\to[6]$, and all $6^r$ such maps are equally likely. Out of these maps $6!\left\{\matrix{r \cr 6\cr}\right\}$ are surjective, where $\left\{\matrix{r \cr 6\cr}\right\}$ (called a Stirling number of the second kind) denotes the number of ways to partition $[r]$ into $6$ nonempty blocks. Therefore the probability $q_r$ that on the $r$ red balls drawn all $6$ numbers are present, is given by
$$q_r={6!\left\{\matrix{r \cr 6\cr}\right\}\over 6^r}\qquad(6\leq r\leq10)\ .$$
It follows that the overall probability $P$ of the event described in the question is given by
$$P=\sum_{r=6}^{10} p_r\>q_r={416216045\over8463329722368}\doteq0.0000491788\ .$$
I'm trying to figure out what the probability of rolling a dice and getting exactly two of the same numbers after we throw the dice $n$ times is?
Exactly two of a specified number, or exactly two of any number?
Case 1 : The probability of getting exactly two sixes. $n\in \{2,3,....\}$
$$\mathsf P(T_6) = {n\choose 2} 5^{n-1}6^{-n} = \frac{5^{n-2}n(n-1)}{2\cdot 6^n}$$
Similarly, $\mathsf P(T_1)=\mathsf P(T_2) = \cdots = \mathsf P(T_6)$
Case 2 : The probability of getting any number exactly twice. $n\in \{2,3, ...\}$
We need a rather complicated inclusion and exclusion, depending on the size of $n$.
$\begin{align}
\mathsf P(\bigcup_{k=1}^6 T_k) & = \sum_{k=1}^6\mathsf P(T_k) - \sum_{k=1}^5\sum_{h=k+1}^6\mathsf P(T_k\cap T_h) + \sum_{k=1}^4\sum_{h=k+1}^5\sum_{j=h+1}^6 \mathsf P(T_k\cap T_h\cap T_j) - \cdots
\\[4ex] & =
6 {n\choose 2, n-2}\frac{5^{n-2}}{6^n} \operatorname{\bf 1}_{\{2,..,\infty\}}(n)
- 15 {n\choose 2,2,n-4}\frac{4^{n-4}}{6^n} \operatorname{\bf 1}_{\{4,..,\infty\}}(n)
+ 20{n\choose 2,2,2,n-6}\frac{3^{n-6}}{6^n} \operatorname{\bf 1}_{\{6,..,\infty\}}(n)
- 15 {n\choose 2,2,2,2,n-8}\frac{2^{n-8}}{6^n} \operatorname{\bf 1}_{\{8,..,\infty\}}(n)
+ 6{n\choose 2,2,2,2,2,n-10}\frac 1{6^n} \operatorname{\bf 1}_{\{10,..,\infty\}}(n)
- {n\choose 2,2,2,2,2,2}\frac 1{6^n} \operatorname{\bf 1}_{\{12\}}(n)
\\[4ex] & =
\frac{6\cdot 5^{n-2}n(n-1)}{2\cdot 6^n} \operatorname{\bf 1}_{\{2,..,\infty\}}(n)
- \frac{15\cdot 4^{n-4}n(n-1)(n-2)(n-3)}{4\cdot 6^n} \operatorname{\bf 1}_{\{4,..,\infty\}}(n)
+ \frac{20\cdot 3^{n-6}n(n-1)(n-2)\cdots(n-5)}{8\cdot 6^n} \operatorname{\bf 1}_{\{6,..,\infty\}}(n)
- \frac{15\cdot 2^{n-8}n(n-1)\cdots(n-7)}{16\cdot 6^n} \operatorname{\bf 1}_{\{8,..,\infty\}}(n)
+ \frac {6\cdot n(n-1)\cdots(n-9)}{32\cdot 6^n} \operatorname{\bf 1}_{\{10,..,\infty\}}(n)
- \frac {12!}{64\cdot 6^{12}} \operatorname{\bf 1}_{\{12\}}(n)
\end{align}$
Best Answer
As you observed, the probability of obtaining doubles on any one throw of a pair of dice is $1/6$ since $6$ of the $36$ possible outcomes result in doubles. Since the probability of obtaining doubles is the same for each of the six trials, this is a binomial distribution problem.
The probability of obtaining exactly $k$ successes in $n$ trials is $$\Pr(X = k) = \binom{n}{k}p^k(1 - p)^{n - k}$$ where $p^k$ is the probability of $k$ successes, $(1 - p)^{n - k}$ is the probability of $n - k$ failures, and $\binom{n}{k}$ counts the number of ways in which exactly $k$ of the $n$ trials could result in a success.
Here, we will define a "success" as a double, so $p = 1/6$. Since there are six trials, $n = 6$. Hence, the probability to get exactly three doubles in six trials is $$\Pr(X = 3) = \binom{6}{3}\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^3$$ In your attempt, you did not take into account the number of ways exactly three of the rolls could result in a success.
If you were instead interested in the probability of at least three doubles in six trials, you would add the probabilities of three or more successes. $$\Pr(X \geq 3) = \sum_{k = 3}^{6} \binom{6}{k}\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{6 - k}$$