# Probablity of rolling the same number twice after n times

Im 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?

The chance to get some number is 1/6 and getting that same number two times in a row must be 1/36, but with this calculation we don't take into account that we throw the dice n times which would increase the probablity that we would get those two numbers exactly two times during n dice throws i'd assume.

What method is used to calculate the probability of such task?

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}

• case 1 was what i meant, cheers.
– John
Commented Nov 19, 2014 at 2:36
• just out of curiousity, isn't the binomial distribution theorem the answer as well? c(n,r)*p^r*q^(n-r) or is there a difference?
– John
Commented Nov 28, 2014 at 12:51