How many times would you have to roll a single die on average to reach a sum of at least 30?
A standard way to solve this is to consider simultaneously the mean number of rolls $t_n$ needed to reach $n$ for every nonnegative integer $n$ (and to remember at the end that the OP is asking for $t_{30}$). So, let us do that...
For every $n\leqslant0$, $t_n=0$. For every $n\geqslant1$, considering the result of the first roll, one sees that $$t_n=1+\frac16\sum_{k=1}^6t_{n-k} $$ Thus, the series $$T(s)=\sum_nt_ns^n$$ solves $$T(s)=\frac s{1-s}+\frac16\sum_{k=1}^6s^kT(s)$$ from which it follows that $$T(s)=\frac{6s}{(1-s)\left(6-\sum\limits_{k=1}^6s^k\right)}=\frac{6s}{6-7s+s^7}$$ To extract the coefficient $t_{30}=[s^{30}]T(s)$ of $s^{30}$ in $T(s)$, rewrite this as $$T(s)=s\left(1-rs\left(1-\frac t7\right)\right)^{-1}=\sum_{n=0}^\infty r^ns^{n+1}\left(1-\frac t7\right)^n$$ where $$r=\frac76\qquad t=s^6$$ hence $$[s^{30}]T(s)=\sum_{k=0}^4r^{5+6k}[t^{4-k}]\left(1-\frac t7\right)^{5+6k}$$ or, equivalently, $$[s^{30}]T(s)=\sum_{k=0}^46^{-5-6k}[t^{4-k}]\left(7-t\right)^{5+6k}=\frac7{6^5}\sum_{k=0}^4(-1)^kr^{6k}{5+6k\choose4-k}$$ that is, $$t_{30}=\frac7{6^5}\left(5-165r^6+136r^{12}-23r^{18}+r^{24}\right) $$ which can be "simplified" into the exact result $$t_{30}= \frac{333366007330230566034343}{36845653286788892983296}$$ with a numerical approximation $$t_{30}\approx 9.047634594384022902065997942672796588425278684184104625$$
Edit: To get some estimates of $t_n$ when $n\to\infty$, note that $$T(s)=\frac s{(1-s)^2Q(s)}$$ where $$Q(s)=\frac16(6+5s+4s^2+3s^3+2s^4+s^5)$$ Furthermore, $(1-s)Q(s)=1-\frac16\sum\limits_{k=1}^6s^k$ has no zero in the closed unit disk except the simple zero $s=1$ hence $Q(s)$ has no zero in the closed unit disk. This implies that $$T(s)=\frac a{(1-s)^2}+\frac b{1-s}+R(s)$$ for some given $(a,b)$ and some rational fraction $R(s)=\sum\limits_nr_ns^n$ with no pole in the closed unit disk. Then, there exists some finite $c$ and some $\varrho$ in $(0,1)$ such that, for every $n$, $$|r_n|\leqslant c\varrho^n$$ This yields, again for every $n$, $$|t_n-a(n+1)-b|=|r_n|\leqslant c\varrho^n$$ Equivalently, $$t_n=an+(a+b)+O(\varrho^n)$$ To identify $(a,b)$, note that $$(1-s)^2T(s)=a+b(1-s)+(1-s)^2R(s)$$ hence $$a=\left.(1-s)^2T(s)\right|_{s=1}=\frac1{Q(1)}$$ and $$b=-\left.\frac d{ds}[(1-s)^2T(s)]\right|_{s=1}=-\frac1{Q(1)}+\frac{Q'(1)}{Q(1)^2}$$ or, equivalently, $$a+b=\frac{Q'(1)}{Q(1)^2}$$ Finally, $Q(1)=\frac72$ and $Q'(1)=\frac{35}6$ hence $a=\frac27$ and $a+b=\frac{10}{21}$, which implies $$t_n=\frac27n+\frac{10}{21}+O(\varrho^n)$$ Edit-edit: More generally, throwing repeatedly a "die" producing a random number of points distributed like $X$, following the same route, one gets $a=E(X)$ and $a+b=Q'(1)/E(X)^2$ with $Q'(1)=\frac12E(X(X-1))$, hence, for some $\varrho_X$ in $(0,1)$ depending on the distribution of $X$, $$t_n=\frac1{E(X)}n+\frac{E(X(X-1))}{2E(X)^2}+O(\varrho_X^n)$$
Let's replace the target "$30$" with a general target $M$. Let $X_n$ denote the 'score' after $n$ throws; define $Y_n = X_n - \tfrac72 n$. Note that $E(Y_n) = 0$, and moreover $E(Y_n \mid Y_{n-1}) = 0$, so it is a martingale. Let $$ \tau = \inf\{ n \ge 0 \mid X_n \ge M \} = \inf\{ n \ge 0 \mid Y_n \ge M - \tfrac72 n \}. $$ Then $\tau$ is a stopping time for $Y$, and so by the optional stopping theorem $E(Y_\tau) = Y_0 = 0$. (Here $\tau$ is deterministically bounded by $M$, and so this is the easy case for OST.) In particular, $$ E(Y_\tau) = 0 \iff E(\tau) = \tfrac27 E(X_\tau). $$ A key point is that while $X_\tau \ge M$, we don't necessarily have equality. Of course, $X_\tau \in [M,M+5]$, and so $$ E(\tau) \in \bigl[ \tfrac27M, \tfrac27(M+5) \bigr], $$ and interval of width $10/7$. If you imagine $M \to \infty$, this means that we know $E(\tau)$ is linear in $M$ (and even know the constant, $\tfrac27$) and our estimate is $\mathcal O(1)$ off.
This is as far as I am able to get. It is not clear to me what the distribution of $X_\tau$ is. (Note that one only needs to know its mean.)
Update. I was interested by this, and posted a very related question: Distribution at First Time a Sum Reaches a Threshold. The answer given there (not by me, unfortunately!) gives the limiting distribution of $X_\tau$, in a simple form from which it is easy to calculate the mean (in the limit $M \to \infty$). Using this question/answer, and a little bit of algebra, one finds that the answer in question is that $$ E(\tau) = \tfrac27 M + \tfrac{10}{21} + o(1). $$
The answer also gives an iterative solution for any $M$, but involves working out a large power of a matrix.