How many times to roll a die before getting $n$ consecutive sixes given $m$ occurences? Given a unbiased dice how to find the average number of rolls required to get $n$ consecutive sixes given already $m$ number of sixes occurred where $m\leq n$. I know how to solve using n consecutive sixes with out any occurrences from this link How many times to roll a die before getting two consecutive sixes?, can anyone help me finding out no of rolls if any occurrences occured before?
 A: For $m\in\left\{ 0,\cdots,n\right\} $ let $\mu_{m}$ denote the expectation
of number of rolls required to get $n$ consecutive sixes on the moment
that the throwing of exactly $m$ consecutive sixes has just occurred. Then $\mu_{n}=0$
and for $m<n$ we find the recursion formula: 
$$\mu_{m}=1+\frac{1}{6}\mu_{m+1}+\frac{5}{6}\mu_{0}$$ It is based on the observation that with probability $\frac{1}{6}$ by throwing a six we land in the situation of having $m+1$ consecutive sixes and with probability $\frac{5}{6}$ we will have to start all over again.
Substituting $m=0$ leads to $\mu_{0}=6+\mu_{1}$ and substituting
this in the recursion formula gives: $\mu_{m}=6+\frac{1}{6}\mu_{m+1}+\frac{5}{6}\mu_{1}$.
Substituting $m=1$ in this new formula leads to $\mu_{1}=6^{2}+\mu_{2}$ and... Now
wait a minute... 
This starts to 'smell' like: $$\mu_{m}=6^{m+1}+\mu_{m+1}$$
doesn't it? Presume this to be true. Then we find: 

$$\mu_{m}=\frac{1}{5}\left(6^{n+1}-6^{m+1}\right)$$

This enables us to actually prove that the presumption is correct. For this it is enough
to check that indeed $\mu_{n}=0$ and $\mu_{m}=1+\frac{1}{6}\mu_{m+1}+\frac{5}{6}\mu_{0}$
for $m<n$, which is just a matter of routine. 
Note that for $n=2$ we find $\mu_{0}=\frac{1}{5}\left(6^{3}-6\right)=42$
corresponding with the answers on the question that you referred to.
