Probability of a letter sequence from a sequence of $n$ random letters What is the probability of this specific sequence in a sequence of $n$ letters -
$A$, then any three letters, then $B$ and then $C$.
My idea: first treat this sequence of 6 bytes as one block. This could occur in $n-6$ places in n letters.
Then, amongst the 6 letters, the number of ways in which three of them are any letters and three are $A, B$ and $C$  specifically is $26^3$.
Then, we want a specific ordering for 3  of these 6 letters. The three bytes in the middle could be in any order, so the number of ways in which to order these 3 occurring  letters is 3!. The other 3 letters $A, B$ and $C$ occur in a specific order. Namely, $A, x , y , z, B, C$.
The total number of combinations of $n$ letters is $26^n$.
Therefore, this probability should be :
$$  \frac{(n-6)(26^3) 3!}{26^n}     $$
Is this correct?
Edit:
@NickPeterson pointed out that $n=6$ will give me a probability of 0.
Thus, I correct my formula to:
$$  \frac{(n-5)(26^3) 3!}{26^n}     $$

This is not an answer.  I am posting this long-winded comment here only for legibility.
Clearly the answer will take the form $$\frac{T}{(26)^n}$$ where $T$ represents the
total # of ways that there is at least one occurrence of the satisfying sequence "A---BC" anywhere in the $n$-character string.
It is dead wrong to compute $T$ as $$(n-5) \times (26)^3.$$
This is because the occurrences of the satisfying sequence "A---BC" in one of the $(n-5)$ slots is not a mutually exclusive event.
This means (for example) that you could have partial strings that look like either "A-A-BCBC" (two occurrences) or "A---BCA---BC" (two occurrences).
One approach is to let $f(k)$ denote the # of ways that exactly $k$ occurrences of the sequence 
"A---BC" can occur.
The idea would then be to compute
$$T = f(1) + f(2) + f(3) + \cdots$$
As has been indicated in the comments, this is very difficult, because the assumption (for example) that the first 6 characters are "A---BC" affects whether another "A---BC" sequence can start in the first 6 characters.
Another approach is to count the # of ways that you can have an $n$-character string where the sequence "A---BC" does not occur.
If I was writing a computer program, I would spin through all 26 characters at each of the $n$ positions.  Then I would discount those situations where the character is a "C", the previous character (if any) is a "B", and the 5th character back (if any) is an "A".
In my opinion, this type of algorithm does no good re manually computing $T$ as a function of $n$.
It is tempting to reason that there are $\binom{n-5}{k}$ ways of distributing k "A"'s among the $(n-5)$ slots.  However, this merely disguises the problem that was broached in the first approach given in this response.  Some of the distributions of k "A"'s will interfere with each other.
I emphasize: this response is not an answer.  The only point to this response is to clarify the challenges in this problem.
