What is the possibility of choosing three balls in a certain sequence given a certain number of attempts? Say I have three balls - red, green and blue in a bag. I will choose a ball from the bag and then return it, then choose another one, etc. Say I will do this 3 times; then the possibility of me picking a certain sequence (for example RGB) is 1/3 * 1/3 * 1/3. But say I do this 4 times. What is the possibility then of running into the same sequence? The desired pick becomes one of the following: BRGB or RRGB or GRGB or RGBB or RGBG or RGBR. I can do the math for this, but what is the general equation where the number of picks is X? Say I want to pick ten balls, what is the possibility of getting RGB then?
For example if I pick four times there is a (1/3)^4 chance that I would get RGBR, but my question isn't the possibility of getting a FOUR letter sequence, rather of getting a three letter sequence in a four letter possibility. The correct answer when x = 4 is (1/3)^4 *6 = 6/81.
Am I explaining this easily? Thanks.
EDIT: Please note that there is only one trial.
 A: Let 


*

*$a(n)$ be the number of strings of length $n$ not containing RGB and not ending R or RG 

*$b(n)$ be the number of strings of length $n$ not containing RGB and ending R  

*$c(n)$ be the number of strings of length $n$ not containing RGB and ending RG  

*$d(n)$ be the number of strings of length $n$ containing RGB 
then starting at $a(0)=1$ and $b(0)=c(0)=d(0)=0$ you have 


*

*$a(n+1)=2a(n)+b(n)+c(n)$

*$b(n+1)=a(n)+b(n)+c(n)$

*$c(n+1)=b(n)$

*$d(n+1)=c(n)+3d(n)$
and $a(n)+b(n)+c(n)+d(n)=3^n$.  You want the probability $d(n)/3^n$.
That is enough to do the calculation, and for ten balls gives $\frac{16293}{59049}\approx 0.2759$ though you could instead use the generating function $\dfrac{x^3}{(1-x)(27-27x+x^3)}.$ 
A: Let $a_n$ be the number of strings of length $n$ that do not contain the sequence RGB; I’ll call these the bad sequences. Clearly $a_0=1,a_1=3$, and $a_2=9$. Suppose now that $n\ge 2$. To build a bad sequence of length $n+1$, to a first approximation you can add any color to the end of a bad string of length $n$; that gives you $3a_n$ strings. However, if the string of length $n$ ends in RG, you cannot add a B. Each bad string of length $n$ ending in RG is the result of adding RG to a arbitrary bad string of length $n-2$, so there are $a_{n-2}$ such bad strings of length $n$ to which we cannot add B. Thus, $a_{n+1}=3a_n-a_{n-2}$. 
Now let $p_n$ be the probability of drawing a bad string of length $n$; $p_n=\dfrac{a_n}{3^n}$, so
$$p_{n+1}=\frac{3a_n-a_{n-2}}{3^{n+1}}=\frac{3^{n+1}p_n-3^{n-2}p_{n-2}}{3^{n+1}}=p_n-\frac{p_{n-2}}{27}\;,$$
and the probability of drawing a string of length $n$ containing RGB is $1-p_n$. The generating function for $a_n$ involves a somewhat intractable cubic, but the recurrences for $a_n$ and $p_n$ aren’t bad.
A: I think you want this:
No. of possibilities with RGB =  $(x-2)3^{x-3}$
$$p = \frac{(x-2)3^{x-3}}{3^{x}} = (x-2)3^{-3} = \frac{x-2}{27}$$
for x=5 you get 27 possibilities, not 24.
However, this only works for $x<6$ as you could get two triples otherwise.
I don't exactly know how avoid this at the moment.

Wrong:
$$X \sim B\left (4,\frac{1}{3}  \right )$$
$$Pr(X \leq 4) = \sum_{i=1}^{4}\binom{4}{i}\left (\frac{1}{9}  \right )^{i}\left (1-\frac{1}{9}  \right )^{i} \approx 0.45753$$
A: Here is an approach that does not count strings:
There are three nonterminal states in this game, namely $s_1$: only the letter $B$ is missing, $s_2$: the two letters GB are missing, and $s_3$: everything else.  Denote by $p_{r,k}$ the probability that we fail when $r\geq0$ picks are left over and we are in state $s_k$. Then $p_{0,k}=1$ $\ (1\leq k\leq 3)$.
The $p_{r,k}$ satisfy the following recurrence:
$$\eqalign{p_{r,1}&={1\over3}p_{r-1,2}+{1\over3}p_{r-1,3}\cr
p_{r,2}&={1\over3}p_{r-1,2}+{1\over3}p_{r-1,1}+{1\over3}p_{r-1,3}\cr
p_{r,3}&={1\over3}p_{r-1,2}+{2\over3}p_{r-1,3}\ .\cr}$$
It follows that ${\bf p}_r={1\over3^r}A^r{\bf 1}$, where $A$ is the matrix
$$A=\left[\matrix{0&1&1\cr 1&1&1\cr 0&1&2\cr}\right]\ .$$
Unfortunately this matrix has unfriendly eigenvalues, so that no simple formula for $p_{n,3}$ results, albeit this number is, of course, rational.
