Expected number of times until getting two 6's What is the expected number of times we need to roll a die until we get two consecutive 6's?
By definition, it is $\sum_{i=1}^\infty i\cdot Pr[X=i]$. If we need $i$ rolls, that means the last two rolls are 6's. But how do we compute the probability that no two consecutive 6's occur before that?
 A: It is reasonably clear that the required expectation exists. Let us call it $a$. Let $b$ be the expected number of additional rolls we need, given that we have not yet met our goal, but have just tossed a $6$.
If the first roll is not a $6$, then we have used $1$ roll, and our conditional expectation, given this happened,  is $1+a$. If the first roll is a $6$, then we have used a roll, and the conditional expectation is $1+b$. It follows that
$$a=\frac{5}{6}(1+a)+\frac{1}{6}(1+b).\tag{1}$$
Suppose now that we have just rolled a $6$, and have not yet met our goal. With probability $\frac{1}{6}$, we roll a $6$. We have used $1$ roll, and the game is over.  With probability $\frac{5}{6}$, we roll a non-$6$, we have used $1$ toss, and the conditional expectation is $1+a$. It follows that
$$b=\frac{1}{6}(1)+\frac{5}{6}(1+a).\tag{2}$$
We have obtained  two linear equations in the two unknowns $a$ and $b$. Solve for $a$. 
Remark: We have shown how to compute the expectation, and not really answered the question about the probability that $X=i$. For finding the expectation, the probability distribution of $X$ is not the most efficient method. However, it is an interesting problem in itself.
The key calculation that needs to be made is the probability that a sequence of length $n$ ends in a non-$6$, and does not have $2$ consecutive $6$'s. One can get a linear recurrence with constant coefficients for the number of "good" sequences of length $n$, and solve the recurrence in any one of the usual ways. 
A: In answer to "What is the probability of having no two consecutive 6s within $n$ rolls",  you may want to build the discrete time Markov chain corresponding to your process, which state space would be defined as such:


*

*0 : I did not just roll a 6

*1 : I just rolled one consecutive 6

*2 : I just rolled two consecutive 6s


And the transition matrix would be as follows:
$$
A = \begin{bmatrix} 5/6 & 1/6 & 0 \\ 5/6 & 0 & 1/6 \\ 0 & 0 & 1\end{bmatrix}
$$
I defined the last state as absorbant.
You just have to raise $A$ to the $n$th power, and the first row will give you for $n$ rolls:


*

*As the sum of the first and second value, the probability that you never have rolled 2 consecutive 6s.

*As third value, the probability that you have rolled two consecutive 6s at some point in the past.

