Initially you are allowed to throw a standard die $10$ times. Each time you roll and successfully get a $6$, you are allowed to throw $5$ more times. So for example, if you get a $6$ in throws $3$, $7$, $12$ and get no other six-s, then you throw a total of $10 + 3\cdot 5 = 25$ times. Let $N$ be the total random number of throws. What is the expected value $\mathbb{E}[N^2]$?
It is easy to compute $\mathbb{E}[N]$ as $10(1 + 5p + (5p)^2 + \cdots)$ where $p=1/6$ is the probability for getting a $6$. Can we compute in a similar tricky manner Var$(N)$ perhaps?
Let $k_n$ be the number of rolls remaining, after $n$ rolls. You end when $k_n=0.$ $k_0=10.$
Then $k_{n+1}=k_n+4$ if the next roll is a six, otherwise $k_{n+1}=k_n-1.$
Then if $N_i$ is the number of rolls to get from $k_n=i+1$ to $k_m=i,$ where $n,m$ are the first occurrences of $i+1,i$ in $k_0,\dots,$ then the $N_i$ are i.i.d.s.
Your $N=N_9+N_8+\cdots+N_0,$ and you get:
$$E(N^2)=10E(N_0^2)+90E(N_0)^2.$$
Now, $E(N_0)=6,$ so the question is, what is $E(N_0^2)?$
Now, $E(N_0^2)=\frac56\cdot1^2+\frac16E\left((1+N_4+N_3+N_2+N_1+N_0)^2\right)$
So $$E(N_0^2)=1+\frac16\left(2\cdot5 E(N_0)+20E(N_0)^2+5E(N_0^2)\right)$$
Again using $E(N_0)=6,$ we get:
$$E(N_0^2)=6+60+720=786.$$
So $$E(N^2)=10\cdot 786+90\cdot 36=11,100.$$
This assume $E(N^2)$ is finite, which might take some proof.
If you replace adding $5$ rolls to adding $6$ or more, you will see $E(N),E(N^2)$ are not finite. So it may require a little work.
If $k_0=k,$ and $r$ is the number of rolls added on six, and $p$ is the probability of a six, you get $E(N)=kE(N_0)$ with: $$E(N_0)=1+prE(N_0).$$ or $E(N_0)=\frac{1}{1-pr}.$ Obviously, this doesn't make sense if $pr\geq 1.$
Now, $$E(N^2)=kE(N_0^2)+(k^2-k)\frac1{(1-pr)^2}$$
And:
$$E(N_0^2)=1+p\left(2rE(N_0)+rE(N_0^2)+(r^2-r)E(N_0)^2\right)$$ or $$ E(N_0^2)=\frac{1}{1-pr}+\frac{2pr}{(1-pr)^2}+\frac{p(r^2-r)}{(1-pr)^3}.$$
If $q=\frac1{1-pr}=E(N_0),$ then:
$$E(N_0^2)=q+2prq^2+(r^2-r)pq^3$$
Then $$E(N^2)=kE(N_0^2)+(k^2-k)q^2=kq(1+2rpq+(k-1)q+(r^2-r)pq^2)$$
If we can find the moment generating function for $N_0,$ $M_{N_0}(t),$ then we can find the moment generating function $M_{N}(t)=M_{N_0}(t)^k.$ Then we can find all the moments, $E(N^d)=M_{N}^{(d)}(0).$
We have:
$$M_{N_0}(t)=E(e^{N_0t})=(1-p)e^t+pM_{N_0}^r(t).$$
So $\alpha=M_{N_0}(t)$ is a root if $p\alpha^r-\alpha+(1-p)e^t=0.$ It seems likely that there are two real roots, though, so that doesn't fully specify $M_{N_0}.$