Let $X$ be the random variable that counts the number of dice throws (fair and 6-sided) till we throw two consecutive $6$. For example, taking the rolls w = $(1,4,3,6,6) \ \ X(w) = 3$. I am trying to find the probability of $P(X=k)$ for all $k$. I know how to do this by using Markov-Chains but i am trying to generalize and find a pattern without using those.
My work so far:
Use uniform probability and use a recursion similar to Fibonacci numbers(for example it would be exactly Fibonacci numbers if you have a coint and you search for the number of throws till you throw TT): start to work out the number of possible strings of dice throws from the end to the start. You know that if X outputs $k$ then the two $6$ appeared in the $(k+1)$th and $(k+2)$th positions and so there must have been a number $\omega(k)\in\{1,...,5\}$ in the $k$th positions (so you have five possibilities to build a string by adding this element. If you keep on with this reasoning for example $\omega(k-1)\in\{1,...,6\}$ (so you have 6 possibilities by adding this element). Now you have to differ between 2 cases for $\omega(k-2)$ :you have $6$ possibilities if $\omega(k-1)\in\{1,...,5\}$ or you have $5$ possibilities if $\omega(k-1)=6$. So the total number of new possibilities is ...