Probability of rolling a die I roll a die until it comes up $6$ and add up the numbers of spots I see. 
For example, I roll $4,1,3,5,6$ and record the number $4+1+3+5+6=19$. Call this sum $S$. 
Find the standard deviation of $S$.

I have been looking for an easy way to do this because I know I can use the definitions here to calculate the variance of $S$ and then take the square root of it. But I am sure there is an easier way to do this.
 A: 
I know I can use the definitions here to calculate the variance of $S$ and then take the square root of it.

Not sure I understand what you mean by that... but here we go. 
Let $i=6$. For every $n\geqslant1$, call $X_n$ the result of the $n$th throw, uniformly distributed on $\{1,2,\ldots,i\}$, and $A_n$ the event that $X_k\ne i$ for every $1\leqslant k\leqslant n-1$. Then 
$$
S=\sum\limits_{n=1}^{+\infty}X_n\cdot[A_n].
$$
For every $n$, $\mathrm E(X_n)=x$ with $x=\frac12(i+1)$ and $\mathrm P(A_n)=a^{n-1}$ with $a=\mathrm P(X_1\ne i)$ hence $a=(i-1)/i$ and
$$
\mathrm E(S)=\sum\limits_{n=1}^{+\infty}x\cdot a^{n-1}=x\cdot(1-a)^{-1}=\tfrac12i(i+1).
$$
Likewise $\mathrm E(S^2)=u+2v$ with
$$
u=\sum\limits_{n=1}^{+\infty}\mathrm E(X_n^2\cdot[A_n]),
\qquad
v=\sum\limits_{n=1}^{+\infty}\sum\limits_{k=n+1}^{+\infty}\mathrm E(X_nX_k\cdot[A_k]).
$$
For every $n$, $\mathrm E(X_n^2)=y$ with $y=\mathrm E(X_1^2)=\frac16(i+1)(2i+1)$, and $X_n$ and $A_n$ are independent, hence
$$
u=\sum\limits_{n=1}^{+\infty}y\cdot a^{n-1}=y\cdot(1-a)^{-1}=yi.
$$
Likewise, for every $k\gt n$, $X_k$ is independent on $X_n\cdot[A_k]$ and 
$$
\mathrm E(X_n\mid A_k)=\mathrm E(X_n\mid X_n\ne i)=z,
$$ 
with $z=\mathrm E(X_1\mid X_1\ne i)=\frac12i$, hence
$$
v=\sum\limits_{n=1}^{+\infty}\sum\limits_{k=n+1}^{+\infty}xz\cdot a^{k-1}=\sum\limits_{n=1}^{+\infty}xz\cdot a^{n}\cdot(1-a)^{-1}=xz\cdot a\cdot(1-a)^{-2}=xzi(i-1).
$$
Finally,
$$
\mbox{Var}(S)=u+2v-\mathrm E(S)^2=yi+2xzi(i-1)-x^2i^2=\tfrac1{12}i(i+1)(i-1)(3i-2).
$$
For $i=6$, $\mathrm E(S)=21$ and $\mbox{Var}(S)=280$.
Edit One sees that $\mathrm E(S)=\mathrm E(X_1)\mathrm E(N)$ where $N$ is the time of the first occurrence of $i$. This is Wald's formula. According to this WP page, the formula for the variance is known as Blackwell–Girshick equation. Proceeding as above, one gets
$$
\mathrm{Var}(S)=\mathrm{Var}(X_1)\cdot\mathrm E(N)+\mathrm E(X_1)^2\cdot\mathrm E(N).
$$
A: Let $Y$ be the number of rolls before a $6$ is rolled. Let $X$ be the sum of the dice rolled before a $6$. Straightforward calculation yields
$$
\mathsf{E}(X|Y=n)=n\;\mathsf{E}(X|Y=1)=3n
$$
and
$$
\mathsf{Var}(X|Y=n)=n\;\mathsf{Var}(X|Y=1)=2n
$$
and
$$
\mathsf{P}(Y=n)=\left(\frac{5}{6}\right)^n\frac{1}{6}
$$
Using the Law of Total Variance, we get
$$
\begin{align}
\mathsf{Var}(X)
&=\mathsf{E}(\mathsf{Var}(X|Y))+\mathsf{Var}(\mathsf{E}(X|Y))\\
&=\sum_{n=0}^\infty\;2n\left(\frac{5}{6}\right)^n\frac{1}{6}+\sum_{n=0}^\infty\;(3n)^2\left(\frac{5}{6}\right)^n\frac{1}{6}-\left(\sum_{n=0}^\infty\;3n\left(\frac{5}{6}\right)^n\frac{1}{6}\right)^2\\
&=\frac{2}{6}\frac{\frac{5}{6}}{(1-\frac{5}{6})^2}+\frac{3^2}{6}\left(\frac{2\left(\frac{5}{6}\right)^2}{(1-\frac{5}{6})^3}+\frac{\frac{5}{6}}{(1-\frac{5}{6})^2}\right)-\left(\frac{3}{6}\frac{\frac{5}{6}}{(1-\frac{5}{6})^2}\right)^2\\
&=10+495-225\\
&=280
\end{align}
$$
Therefore, the standard deviation is $\sqrt{280}$.
Afterword:
Although not requested in the question, the expected value of $S$ is simple to compute by the linearity of expectation. Since the probability of rolling a $6$ is $\frac{1}{6}$, the mean number of rolls is $6$. Since each non-$6$ roll has a mean of $3$ and on average there will be $5$ non-$6$ rolls, we get $\mathsf{E}(S)=5\cdot3+6=21$.
We can also compute this using the set-up for the variance above. Since $S=6+X$,
$$
\begin{align}
\mathsf{E}(S)
&=6+\mathsf{E}(X)\\
&=6+\mathsf{E}(\mathsf{E}(X|Y))\\
&=6+\sum_{n=0}^\infty3n\left(\frac{5}{6}\right)^n\frac{1}{6}\\
&=6+\frac{3}{6}\frac{\frac{5}{6}}{(1-\frac{5}{6})^2}\\
&=6+15\\
&=21
\end{align}
$$
A: A considerably less general and less detailed solution than the one provided by Didier Piau is as follows.
If the first $6$ occurs on the $N$-th roll of the die, then $N$ is a geometric random variable with parameter $\frac{1}{6}$.  We have that $E[N] = 6$ and 
$\text{var}(N) = 30$.  Given the value of $N$, we can write
$$S = 6 + \sum_{i=1}^{N-1} Y_i$$ where the $Y_i$ are independent  random 
variables uniformly distributed on $\{1,2,3,4,5\}$ and the $6$ is the
contribution of the $N$th roll of the die to the sum. 
Since  $E[Y_i] = \frac{1+2+3+4+5}{5} = 3$,
$$E[S\mid N] = E\left[6 + \sum_{i=1}^{N-1} Y_i\right]
= 6 + \sum_{i=1}^{N-1}E[Y_i] = 6 + 3(N-1) = 3N + 3$$
and so 
$$E[S] = E[E[S|N]] = E[3N + 3] = 21.$$
Since each $Y_i$ has variance $\dfrac{1^2+2^2+3^2+4^2+5^2}{5}-3^2 = 2$ and 
is independent of the
others, 
$$\text{var}(S\mid N) 
= \text{var}\left(6 + \sum_{i=1}^{N-1} Y_i\right) 
= \sum_{i=1}^{N-1}\text{var}(Y_i) = 2(N-1) ~$$
and so
$$E[\text{var}(S\mid N)] = E[2(N-1)] = 10$$ 
while
$$\text{var}(E[S\mid N]) = \text{var}(3N + 3) = 9\cdot\text{var}(N) = 270$$
giving
$$\text{var}(S) = E[\text{var}(S\mid N)] + \text{var}(E[S\mid N]) = 280.$$
