(Scenario with 1 dice):
Denote $P(r,s)$ probability to get sum $s$ by $r$ rolls.
Denote $Q(r,s) = 1-\sum\limits_{q=1}^{s-1}P(r,q)$ $~$ probability to get sum $\ge s$ by $r$ rolls.
Obviously:
$$
P(1,1)=P(1,2)=\ldots=P(1,10)=0.1;
$$
then
$$
P(r,s) = \sum\limits_{q=s-10}^{s-1} P(r-1,q)\cdot 0.1
$$
(i.e. sum $s$ we can get of "previous" sums $s-10$, or $s-9$, ..., or $s-1$).
Step-by-step, we can fill table of probabilities $P(r,s)$.
A few conclusions:
$Q(16,100) = 1-\sum\limits_{q=1}^{99}P(16,q) \approx 0.159924$;
$Q(17,100) \approx 0.307628$;
$Q(18,100) \approx 0.483771$;
$Q(19,100) \approx 0.654096$;
$Q(20,100) \approx 0.791809$;
$Q(21,100) \approx 0.887108$;
$Q(22,100) \approx 0.944590$;
$Q(23,100) \approx 0.975252$.
Now you can choose closest values:
$25 \%$: ~ 17 rolls;
$50 \%$: ~ 18 rolls;
$75 \%$: ~ 20 rolls;
$90 \%$: ~ 21 rolls.
(Scenario with 2 dices):
As dices are independent, then all that we need is to consider even number of single rolls, and divide them by $2$.
$Q_2(8,100) =Q(16,100) \approx 0.159924 (\approx 15.99 \%);$
$Q_2(9,100) =Q(18,100) \approx 0.483771 (\approx 48.38 \%);$
$Q_2(10,100)=Q(20,100) \approx 0.791809 (\approx 79.18 \%);$
$Q_2(11,100)=Q(22,100) \approx 0.944590 (\approx 94.46 \%).$
As you are guranteed to get a repeat after at most seven rolls, this is a very finite problem and we can simply enumerate all cases.
Here are possible runs of the consecutive dice rolls, where distinct letters stand for distinct numbers coming up, and the rolling is stopped as soon as we have a repeat.
$aa$: probability $\frac16$ (need the second roll to be the same as the first)
$aba$ or $abb$: probability $\frac56 \cdot \frac26$ (second roll distinct from first, and third roll same as any of the first two)
$abc$ and then repeat (that is, $abca$ or $abcb$ or $abcc$): $\frac56 \frac46 \cdot \frac36$
$abcd$ and then repeat: $\frac56 \frac46 \frac36 \cdot \frac46$
$abcde$ and then repeat: $\frac56 \frac46 \frac36 \frac26 \cdot \frac56$
$abcdef$ and then repeat: $\frac56 \frac46 \frac36 \frac26 \frac16 \cdot 1$
Thus, the number of times a dice is rolled before seeing a repeat is $2$ with probability $\frac16$, $3$ with probability $\frac5{18}$, and so on. In general, the probability that we roll $k+1$ times before seeing a repeat is
$$\frac{n}{n}\frac{n-1}{n}\frac{n-2}{n}\cdots\frac{n-k+1}{n} \frac{k}{n} = \frac{n!k}{(n-k)!n^{k+1}}$$
From these numbers you can calculate the expected value, the thresholds for 50% or 70% certainty, etc.
For $n=6$, doing these calculations gives the expected number of times you roll the dice until seeing a repeat to be (using $r$ below for $k+1$, the number of rolls):
$$\operatorname{E}[r] = \sum_{r=2}^7 r \frac{6!(r-1)}{(6-r+1)!6^r} = \frac{1223}{324} \approx 3.77$$
The cumulative probability distribution looks like the below:
Thus for example, you need to roll the dice at most $5$ times, with probability about $90\%$, at most $4$ times with probability about $70\%$, etc.
Best Answer
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 $X$, $$t_n=\frac1{E(X)}n+\frac{E(X(X-1))}{2E(X)^2}+O(\varrho_X^n)$$