Player rolls dice: for every “1” he gets 1 point, “11” - 5 points, “111” - 10 pts, so on. What is the mean score after 100 rolls? For $x \neq 1$:

*

*every $“\cdots x1x\cdots“$ gives +1 pt

*every $“\cdots x11x\cdots“$ gives +5 pts.

*every $“\cdots x111x\cdots“$ gives +10 pts.

*And so on: $n$ consecutive 1’s gives us $(n-1)5$ points.

To make it clear, the usual 6-sided dice is rolled 100 times, so for example if player rolls dice 1 time - there’s 1/6 chance of getting 1 point; rolls two times - then there’s $\frac{1}{6}\frac{5}{6}2$ chance of 1 point(“1x” or “x1”) and $\left(\frac{1}{6}\right)^2$ of getting 5 points(only if “11” rolled out).
The question: what is the mean score after rolling dice 100 times?
The problem is: how do we calculate the mean when the number of rolls is so huge? It’s clear that using definition of the mean directly is not an option because number of different configurations for getting any score is immense(only if that score is not, say, 99*5 which require all the 1’s).
I tried to use induction, but it didn’t worked out, for 3-4 rolls it already gets complicated. Moreover, how am I suppose to use it? If I know mean for &n& rolls and then I add $(n+1)$th roll - it will add 0, 1 or 5 points depending on which number rolled in $n$th place. Seems like knowing mean for $n$ rolls won’t be much of a help because after one more roll chance of getting any score is different.
Another idea given to me by roommate is to fix number of ones that we get in entire 100-length sequence(so probability is fixed as well), and see what number of points we can possibly get with that number of 1’s - to know that these numbers will appear in formula for mean with known probability factor. But I’m not sure about that also because the amount of combinations is still insane.
I ran out of ideas for now. Feels like there must be some efficient, less bloody way to calculate all that because our teacher gave us only 40 minutes for that problem (and another one), which completely freaked me out. All I wanted to say - I really appreciate any of your help since I absolutely have to figure this out.
One more question: could anyone recommend some book with hard combinatorial problems in probability? Or some good textbook which could explain how to solve problems of that kind. That would be very helpful as well, thank you.
 A: Just to give a different approach, we could use indicator variables to count the expected occurrences of blocks of exactly $n$ ones.
We note, for instance, that the expected number of singleton $1's$ is $$E_1=2\times \frac 16\times \frac 56+98\times \frac 16\times \left(\frac 56\right)^2$$
Where the first term counts the contribution from the first and last tosses and the second term counts the contribution for all the middle terms.  Note that blocks in the middle must be preceded and followed by something other than $1$.
Similarly, the expected number of blocks of exactly $n$ ones is $$E_n=2\times \left(\frac 16\right)^n\times \frac 56+(99-n)\times \left(\frac 16\right)^n\times \left(\frac 56\right)^2$$
At least for $2≤n≤99$.   For $100$ there's only the one possibility and we get $E_{100}=\left(\frac 16\right)^{100}$.
It follows that the answer is $$E_1\times 1 +\sum_{n=2}^{100}E_n\times 5(n-1)\approx 25.3704$$
A: Let $a_n$ be the expected score of $n$ rolls. Obviously $a_0 = 0$.
Let $k$ be the number of consecutive $1$'s at the beginning of the rolls. E.g. $k=0$ if the first roll is not $1$.
The probability of $k$ consecutive $1$'s at the beginning is equal to $5/6^{k+1}$ for $0\leq k <n$ and equal to $1/6^n$ for $k=n$.
In that case, the remaining rolls will give an expected score of $a_{n-k-1}$ (where $a_{-1}$ is understood to be $0$).
Thus we get the recurrence relation:$$a_n = \frac1 {6^n}(5(n-1)+1_{n=1})+\sum _{k=0}^{n-1}\frac 5 {6^{k+1}}(a_{n-k-1} + 5(k-1) + 1_{k=1} + 5\cdot 1_{k =0})$$ for all $n\geq1$.
It is then easy to show by induction that $a_n = (55n-20)/216$ for all $n\geq2$.
Therefore the answer for $100$ rolls is $685/27\approx
25.37$.
A: Seems like I found a solution.
Since contribution of each roll is dependent on its neighbours, and the mean function is linear operator which doesn’t care about dependence - let’s assign random value to each of 100 rolls, making their sum equal to overall score.
$n_i$ denotes i-th roll; $x$ is anything that’s not 1;
Defining value $\xi_1$ which we gonna assign to 1st and 100th roll:
$$\begin{array}{c|c|} 
\text{$n_1n_2$} & \text{$\xi_1$} \\ \hline
xx, x1 & 0 \\ \hline
1x & 1 \\ \hline
11 & 4 \\ \hline
\end{array}$$
In other words, it denotes how much 1st roll adds to the sum. In case of $n_1n_2n_3 = 112, 113, \cdots$, $n_1$ gives +4 and $n_2$ gives +1.
Next we define $\xi_i$, $i = 2, 3, \cdots, 99$
$$\begin{array}{c|c|} 
\text{$n_{i-1}n_{i}n_{i+1}$} & \text{$\xi_i$} \\ \hline
1x1, xx1, xxx, 1xx & 0 \\ \hline
x1x, 11x & 1 \\ \hline
x11 & 4 \\ \hline
111 & 5 \\ \hline
\end{array}$$
$S$ - score.
$$S = \xi_1 + \cdots + \xi_{100} ~ ,$$ $$ \mathbb{E}S = \mathbb{E}\xi_1 + \cdots + \mathbb{E}\xi_{100} = 2\mathbb{E}\xi_1 + 98 \mathbb{E}\xi_2 ~ ,$$
If I’m not messed up here, $\mathbb{E}\xi_1 = 1/4$ and $\mathbb{E}\xi_2 = 55/216$, so the answer is
$\frac{1}{2} + 98\frac{55}{216} \approx 25,45$
(wow, which even correlates with QC_QAOA’s machinery answer in the comments)
A: This is not an answer, but simply me checking what happens for smaller values of rolls. Note that these are exact values and not estimations as I am going through all $6^n$ different roll possibilities for $n$ rolls.
$$n=1:\ 1\cdot 6^{-1}$$
$$n=2:\ 15\cdot 6^{-2}$$
$$n=4:\ 145\cdot 6^{-3}$$
$$n=4:\ 1200\cdot 6^{-4}$$
$$n=5:\ 9180\cdot 6^{-5}$$
$$n=6:\ 66960\cdot 6^{-6}$$
$$n=7:\ 473040\cdot 6^{-7}$$
$$n=8:\ 3265920\cdot 6^{-8}$$
This tracks with @WhatsUp answer as it matches their function for the first few values.
A: This problem would be easier if the rule of $5(n-1)$ held also for $n=1$; that is, if an isolated $1$ gave no points instead of $1$ point.  That leads to the insight of solving the easier problem, then adding the expectation of points due to isolated $1$s.
[This is similar to the approach taken by @lulu.]
For the sequences, each $1$ will contribute $5$ points if and only if the previous roll was a $1$ as well.  (We can consider that the first $1$ in a sequence contributes nothing, other than enabling the next roll to contribute if it is also a $1$.) Since the first roll can't contribute this way, each of dice $2$ through $100$ -- $99$ dice -- contributes an expectation of $5$ points $\times \frac16$ (this roll is a $1$) $ \times$ $\frac16$ (previous roll was a $1$) for a total of $\frac{495}{36}$.
For the isolated $1$s, the $98$ dice that have two neighbors each contribute $\frac16$ (this roll is a $1$) $\times \frac{25}{36}$ (neither neighbor was a $1$) for a total of $\frac{98\cdot 25}{216}$.  The two endpoint dice each contribute $\frac15\times\frac56$ for an additional $\frac{10}{6}$.
The total expectation is
$$\frac{495}{36}+ \frac{2450}{216} + \frac{10}{6} = \frac{2970 + 2450 + 360}{216} = \frac{1445}{54} \approx 26.76$$
I realize this answer disagrees with the previous answers.  The approach is simple enough that I have confidence that this answer is correct.
