Yes, there is a way to get the exact answer.
Definition & Notation:
$X_i$: number of throw to get the same number (let say 6) in $i$ consecutive rolls,
$\mu_i = E(X_i)$, that is the expected number of roll; $\mu_6$ is what we want.
$D$: Indicator of whether get the number we desire (6 in this case) in one throw, 1 for success, 0 for fail. It is clearly that $D \sim Bernoulli(p)$.
$p$: the probability of getting the number (6) we desire
Let's us start from the simplest case $i=1$, the number of throw to get one 6. Some of you may quickly realised that
$$\mu_1 = \frac{1}{p}$$
In fact, when $i=1$, we are doing a sequence of independent bernoulli trial until we make the first success (get one 6) and $X_1$ is the number of trial needed until the first, which follows a geometric distribution$^1$$^2$$^3$ with parameter $p$, and its mean is $\frac{1}{p}$.
So, how about $i\ge2$? In order to get r consecutive 6's in die rolling, we must get (r-1) consecutive 6's first. At that moment we have 2 scenarios:
- We get a 6 in the next throw and end
- We don't get a '6' and keep playing with our die until we get r consecutive 6's
As you will see, the number of throw is $X_{r-1}+1$ in the first scenario, and $X_{r}+1+X_{r}$ in another. Don't forget that you have the count the throw right after you get (r-1)-time 6's, so we have '+1' when we calculate both counts. Combining them, we have
$$X_r = X_{r-1} + 1 + (1-D)X_r$$
Using double expectation:
$$\mu_r = E(X_r) = E\left[E(X_r|D)\right]$$
$$\implies \mu_r = pE\left[E(X_r|D=1)\right] + (1-p)E\left[E(X_r|D=0)\right]$$
Since
$$(X_r|D=1) = X_{r-1} + 1 \implies E(X_r|D=1) = \mu_{r-1} + 1$$
$$(X_r|D=0) = X_{r-1} + 1 + X_r \implies E(X_r|D=0) = \mu_{r-1} + 1 + \mu_r$$
we have
$$\mu_r = p[\mu_{r-1}+1] + (1-p)[\mu_{r-1} + 1 + \mu_r]$$
After some rearrangement, we will get the following relationship:
$$\mu_r = \frac{\mu_{r-1}+1}{p}$$
If we are playing a fair die (i.e. $p=\frac{1}{6}$), we will obtain $\mu_6 = 55986$.
Now, we relax our criteria, let say any number (either 1, 2, 3, 4, 5, 6) appears six times in a row. The answer is truly smaller than 55986 because you can six 1's before getting six 6's. To calculate the expected number of throw, same logic used above still apply:
$Y_i$: number of throw to get the same number (either 1, 2, 3, 4, 5, 6) in $i$ consecutive rolls,
$\theta_i = E(Y_i)$, that is the expected number of roll; $\theta_6$ is what we want.
$D_x$: Indicator of getting $x$, $x\in\{1,2,3,4,5,6\}$.
First, we have $\theta_1=1$ this time. The reason is obvious, when we throw a die, we must get a number from {1, 2, 3, 4, 5, 6} and end rolling the die afterwards, the count ($X_1$) must be 1, also $\mu_1$, unless you are not playing a 6-sided die or the die disappeared after you throw it etc.
And for $r \ge 2$, let's go back to the scenario of getting $x$ (r-1)-time
- We get a $x$ in the next throw and end
- We don't get a $x$ but $y$, $y \ne x$, keep playing with our die until we get r consecutive $y$
For scenario 1, only ($Y_{r-1}+1$) rolls are needed, but for another scenario, we need ($Y_{r-1}+Y_r$). Be noted that no '+1' is needed here because when a different number $y$ shows up, we are attempt to get $r$ consecutive $y$, the 'first' roll is included in $Y_r$ here. Then, the following can be derived:
$$Y_r = D_x(Y_{r-1}+1) + (1-D_x)(Y_{r-1}+Y_r), x \in \{1,2,3,4,5,6\}$$
If we are playing a FAIR die, the recursive relation can be generalized as
$$Y_r = X_{r-1} + (1-D)Y_r$$
where $D$ is the indicator of getting the same number as last throw given that we are playing a fair die, $Pr(D=1)=\frac{1}{6}$.
After all, the following can be derived,
$$\theta_r = 6\theta_{r-1}+1$$
Applying the formula, $\theta_6 = 9331$ will be obtained under the looser criteria (fair die). (Credit to Aksakal for pointing out my mistake).
For the second question, we can make use of the transition matrix $M$
$$M = \left[\begin{matrix}
\text{stage} & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
0 & \frac{5}{6} & \frac{1}{6} & 0 & 0 & 0 & 0 & 0\\
1 & \frac{5}{6} & 0 & \frac{1}{6} & 0 & 0 & 0 & 0\\
2 & \frac{5}{6} & 0 & 0 & \frac{1}{6} & 0 & 0 & 0\\
3 & \frac{5}{6} & 0 & 0 & 0 & \frac{1}{6} & 0 & 0\\
4 & \frac{5}{6} & 0 & 0 & 0 & 0 & \frac{1}{6} & 0\\
5 & \frac{5}{6} & 0 & 0 & 0 & 0 & 0 & \frac{1}{6}\\
6 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\
\end{matrix}\right]$$
where stage $i$ means we currently have $i$ consecutive '6' and we assume that we are playing a fair die.
The entries of the matrix are the probability of getting from one stage (row) to one stage (column). If we die once; for example, from stage 6 to stage 6, the probability is 1 (because we have achieved our goal)
By the properties of transition matrix, $M^k$ is the transition matrix of rolling the die k-time. Therefore, we can calculate $M^{100}$ and value in the first row and the last column will be the probability of going to stage 6 from stage 0 after 100-time of roll dicing, that is your answer. (it is 0.001699134)
If we loosen our criteria to any number six times in a row, our transition matrix will become
$$M = \left[\begin{matrix}
\text{stage} & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\
1 & 0 & \frac{5}{6}& \frac{1}{6} & 0 & 0 & 0 & 0\\
2 & 0 & \frac{5}{6} & 0 & \frac{1}{6} & 0 & 0 & 0\\
3 & 0 & \frac{5}{6} & 0 & 0 & \frac{1}{6} & 0 & 0\\
4 & 0 & \frac{5}{6} & 0 & 0 & 0 & \frac{1}{6} & 0\\
5 & 0 & \frac{5}{6} & 0 & 0 & 0 & 0 & \frac{1}{6}\\
6 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\
\end{matrix}\right]$$
For question 1 (six consecutive '6'), there is another way of think (which is my original answer) but similar approach. It will yield a slightly different result and the reason is still unknown. Any discussion are all welcome.
Definition & Notation:
stage $i$: the moment that we have $i$ 6 in a row
$X_i$: the random variable of no. of rolls needed in stage $i$ in order to get "6" six times
$\mu_i = E(X_i)$, in this case, $\mu_0$ is our answer.
Suppose we have now diced $n$-times and get five 6's in a row (stage $5$) and roll the die again. If we get a 6, we achieve our goal and the count of rolling is $n+1$; otherwise, we go back to stage $0$ and $(1 + X_0)$-time will be needed to archive our goal. Thus, the expected value of $X_5$ ($\mu_5$) will be
$$\mu_5 = \frac{1}{6} + \frac{5}{6}(1+\mu_0) = 1 + \frac{5}{6}\mu_0$$
Now, consider we have four 6's in a row (stage $4$). If we get a 6 next, we proceed to stage $5$, that means we have to roll $(1+X_5)$-time more; otherwise, we are back to stage $0$. Thus,
$$\mu_4 = 1 + \frac{1}{6}\mu_5 + \frac{5}{6}\mu_0$$
Following the same logic, the followings will be obtained:
$$\mu_3 = 1 + \frac{1}{6}\mu_4 + \frac{5}{6}\mu_0$$
$$\mu_2 = 1 + \frac{1}{6}\mu_3 + \frac{5}{6}\mu_0$$
$$\mu_1 = 1 + \frac{1}{6}\mu_2 + \frac{5}{6}\mu_0$$
$$\mu_0 = 1 + \frac{1}{6}\mu_1 + \frac{5}{6}\mu_0$$
So, we get a system of linear equations:
$$\left[\begin{matrix}
\frac{1}{6} & -\frac{1}{6} & 0 & 0 & 0 & 0 \\
\frac{5}{6} & -1 & \frac{1}{6} & 0 & 0 & 0 \\
\frac{5}{6} & 0 & -1 & \frac{1}{6} & 0 & 0 \\
\frac{5}{6} & 0 & 0 & -1 & \frac{1}{6} & 0 \\
\frac{5}{6} & 0 & 0 & 0 & -1 & \frac{1}{6} \\
\frac{5}{6} & 0 & 0 & 0 & 0 & -1 \\
\end{matrix}\right]
\left[\begin{matrix}\mu_0 \\ \mu_1 \\ \mu_2 \\ \mu_3 \\ \mu_4 \\ \mu_5\end{matrix}\right]
=
\left[\begin{matrix}1\\ -1\\ -1\\ -1\\ -1\\ 1\end{matrix}\right]
\implies
\left[\begin{matrix}\mu_0 \\ \mu_1 \\ \mu_2 \\ \mu_3 \\ \mu_4 \\ \mu_5\end{matrix}\right]
=
\left[\begin{matrix}55974\\ 55968\\ 55932\\ 55716\\ 54420\\ 46644\end{matrix}\right]
$$
