TITLE: Roll 10 fair die. What is the probability that the number 1 appears exactly four times, on four consecutive rolls
QUESTION [5 upvotes]: I am asked:

Roll $10$ fair die. What is the probability that the number $1$ appears
  exactly four times, on four consecutive rolls?

The answer I was given is
$$\frac{1}{6^{10}}\cdot7\cdot5^6$$
with a comment saying we multiply by 7 "for the first roll".
However, this doesn't make sense to me since there are only $6$ possible sides to the die. So it's not counting sides of the die. 
The only reason I can come up with for the 7 appearing is the case when the first $6$ rolls do not produce a sequence of consecutive $1$'s, so the $7$th item must begin the sequence of consecutive $1$'s
So where exactly is this $7$ coming from?

REPLY [5 votes]: For a better understanding, the seven alternative cases are:
 [1 1 1 1] * * * * * *
 * [1 1 1 1] * * * * *
 * * [1 1 1 1] * * * *
 * * * [1 1 1 1] * * *
 * * * * [1 1 1 1] * *
 * * * * * [1 1 1 1] *
 * * * * * * [1 1 1 1]

The six stars are $2,3,4,5,6$ (five categories) and can take $5^6$ combinations. Hence the numerator is $(7)5^6$. The total number of combinations is $6^{10}$.