Is this Approach correct? Determine number of dice rolls required to obtain the 3rd five If we roll a pair of dice repeatedly, how many rolls do we need to obtain the nth five (example 3rd five)?  
I think this can be modeled using a negative binomial distribution. Is this naive?   
If we want to calculate the Pr(N>5) where N is the number of rolls required, can I approach it as 1-Pr(N<=4)?
- If so, does that mean I calculate
    $\sum\limits_{i=0}^4 (negativeBin(i,p))$ ?
 A: Suppose that we perform an experiment independently until the $k$-th success, where the probability of success on any trial is $p$. Let $N$ be the number of trials we use. 
Then $N$ has negative binomial distribution with parameters $k$ and $p$. Let us find the probability that $N=n$.
We have $N=n$ if we have $k-1$ successes in the first $n-1$ trials, and success on the $n$-th trial. Thus
$$\Pr(N=n)=\binom{n-1}{k-1}p^{k-1}(1-p)^{n-k}p=\binom{n-1}{k-1}p^{k}(1-p)^{n-k}.\tag{1}$$
Suppose for example that $k=3$ and we want $\Pr(N\gt 5)$. We have $N\gt 5$ precisely if it is not the case that $N\le 5$. Thus $N\gt 5$ has probability
$$1-(\Pr(N=3)+\Pr(N=4)+\Pr(N=5)).$$
For the above probabilities, use Formula (1), with $p=1/6$.
Remark: The negative binomial $N$ with parameters $k$ and $N$ is the sum of $k$ (independent) geometric random variables with parameter $p$. Thus $E(N)=\frac{k}{p}$.
A: To have Bernoulii success number $n$ on the roll number $k$, you need $n-1$ successes and $k-n$ failures in any order, then one more success, at success probability $1/6$.
$$\mathsf P(X_n=k) = \binom{k-1}{n-1} \frac{5^{k-n}}{6^k}$$
Now find the expectation of this $$\begin{align}
\mathsf E(X_n) & =\sum_{k=n}^\infty k\binom{k-1}{n-1}\frac{5^{k-n}}{6^k}
\\[1ex] & = 6 n
\end{align}$$

Alternatively.  As it is a Geometrically Distributed, we expect the first success on roll $6$.  After obtaining the first success, we then expect the second success on the $6$ roll after the first one, and so on.
Then, because of linearity of expectation, we expect success $n$ on roll $6n$.
