Probability of getting 6 k times in a row What is the probability of getting $6$ $K$ times in a row when rolling a dice N times?
I thought it's $(1/6)^k*(5/6)^{n-k}$ and that times $N-K+1$ since there are $N-K+1$ ways to place an array of consecutive elements to $N$ places.
 A: In the following for convenience I write $p$ for $\frac{1}{6}$ and $q$ for $\frac{5}{6}$
Let $A_{K,N}$ denote the event that among the first $N$ throws there
is a consecutive row of $K$ sixes.
Let it be that the $F$-th throw is the first throw that does not
give a six, so that $P\left(F=r\right)=p^{r-1}q$.
Then $P\left(A_{K,N}\mid F=r\right)=1$ if $r>K$ and $P\left(A_{K,N}\mid F=r\right)=P\left(A_{K,N-r}\right)$
otherwise. 
Further $P\left(A_{K,N-r}\right)=0$ if $K>N-r$.
So for $K\leq N$ we find:
$$P\left(A_{K,N}\right)=\sum_{r=1}^{\infty}P\left(A_{K,N}\mid F=r\right)P\left(F=r\right)=\sum_{r=1}^{K}P\left(A_{K,N-r}\right)p^{r-1}q+p^{K}$$
Based on this equality with induction it can be proved that:
$$P\left(A_{K,N}\right)=p^{K}\left(1+\left(N-K\right)q\right)\text{ for }
N\in\left\{ K,K+1,\dots,2K\right\} $$(which is not valid for $N>2K$).

See also this question that was posed by me once at a time.
A: This is really just a long comment, but I thought that it would be a mess to write it as a comment:
Please clarify your question with, at least, the following data (some of the answers are in the comments):


*

*Does the run of $6$'s need to be the only $6$'s in the string?  If $k=2$, would $6656$ be allowable?

*Can the run of $6$'s be longer than $k$?  If $k=2$, would $6665$ be counted?

*Can you have more than one run of $6$'s in the string?  If $k=2$, would $66566$ be allowable?
A: So, let's say that X is a random variable for tracing the number of 6s.
Let's say that C is the condition for the 6s to be consecutive.
Since the task is to find the probability of gaining at least K subsequent 6s, we can look for probability of event A-at least K subsequent 6s fell this way:
     P(A)=P(X=K|C)+P(X=K+1|C)+...+P(X=K+N|C)+...,
where P(X=K|C) is the probability of K 6s falling under the condition that they are consecutive.
Since the event that K 6s have fallen is independent from the event that all the 6s are in order, we can say that:
    P(X=K|C)=P(X=K)*P(C).
Is this the right way to go?
Edit:
Doesn't seem right because the events ARE independent, because K 6s need to fall in order for them to be consecutive.
