Rolling A Die 10 times How do I find the probability of not rolling the same number exactly 4 times in a row when I roll a die 10 times?
How do I use complementary counting for this if it is the best method?
 A: I would set it up as follows, I think you can fill in the details:
Identify the events of 10 dice rolls with $\Omega := \{1, \dots, 6\}^{10}$, and let $M \subset \Omega$ the subset of rolls with at least one number exactly $4$ times in a row. Then the probability you need is $$P = 1 - \frac{\#M}{6^{10}}$$.
Given $a \in \{1,\dots,6\}$, in an event, $a$ can occur 4 times in a row as follows (for $i \in \{1, \dots, 7\}$):
$$(b_1,\dots, b_{i-1}, a, a, a, a, b_{i+4}, \dots, b_{10})$$
Here you want that $b_{i-1}, b_{i+4} \neq a$, all other $b_j$'s don't matter. 
So for example for $i = 4$ (i.e. the rows of $a$'s start at fourth position), there are 
$$6 \cdot 6 \cdot 5 \cdot 5 \cdot 6 \cdot 6 = 6^4\cdot5^2$$
possibilities.
From here I think you can proceed on your own:


*

*Count the possibilities for any $i \in \{1, \dots, 7\}$ (don't forget there are special cases for $i = 1,2$).

*Keep in mind that there are 6 possibilities for $a$

*Keep in mind that you are counting events twice, where there are two 4-rows. This can happen for two different numbers, as well as for $a$, e.g. $(a,a,a,a,b,c,a,a,a,a) \text{ or } (b,d,d,d,d,a,a,a,a,c)$ 


I hope I didn't miss anything.
A: There is a derivation of the formulation for such problems here. 
http://ruangbacafmipa.staff.ub.ac.id/files/2012/02/An-Introduction-to-probability-Theory-by-William-Feller.pdf
Chapter XIII, section 10. Renewal Theory
The exact formulation is in page 341 section (25) (last equation on that page)
