What is the expected number of rolls until you see a 6 immediately followed by a 2?
Let $D_i$ = number observed on the $i^{th}$ roll
Assume $T_0=0$
Now let $T_1$ = min{$i\ge2$:$D_{i-1}=6$, $D_i=2$}
Hence, for $k=1,2,...$, let $T_k$ = time of the occurence of "6 then 2"
Then the $T_k$'s can be viewed as the renewal times of a renewal process, and all have the same distribution and are independent.
So, $M(n)=E(N(n))$, by definition. (M is the renewal function and N is the number of renewals on an interval)
Then, we have
$M(n)=E(N(n))$
$=\sum_{k=1}^n$ P(there is a renewal on time k)
$=\sum_{k=2}^n$ P($D_{k-1}=6$, $D_k=2$)
=$(n-1)(\frac1{36})$
Q: What do I do to get this last step? I do understand the steps needed to answer the question itself, I'm just not understanding the summation portion.