Probability of rolling 3 sided die X times We have a fair $3$ sided die $(a,b,c)$, and we perform the following experiment:
Roll the die until we have seen $10$ of any of the sides, let $X$ be the number of times the die was rolled.
eg. ${bcbaaaaaaaaaa,bbbbbbbbbb,aaaaaaacacccccccccc}$
We are interested in the probability of $X$ for all possible values of $X$, ie. $X={10,11...28}$.
 A: Let $P(x,A,B,C)$ be the probability that there will be $x$ more rolls when we have already seen the $a$ face $A$ times, the $b$ face $B$ times, and the $c$ face $C$ times. Then 
$$
P(x,A,B,C)=
\begin{cases}
\frac13\left(\begin{array}{l}P(x-1,A+1,B,C)+\\P(x-1,A,B+1,C)+\\P(x-1,A,B,C+1)\end{array}\right) & \text{if }A,B,C<10
\\
1 & \text{if }A=10\text{ or $B=10$ or $C=10$ and $X=0$}
\\
0 & \text{if }A=10\text{ or $B=10$ or $C=10$ and $X\neq0$}
\end{cases}
$$
You want the probabilities $P(x,0,0,0)$ for each $x$. You can easily compute these using a memoized recursive program, for example this one here.
A: Not sure if this is what you have in mind when you said you wanted an "analytical" way...
By symmetry, the final (i.e. $X$th) die-roll is $a$ or $b$ or $c$ with equal probability.  So:


*

*$P(X=m) = 3 \times P(X=m, \text{final roll is } a)$.


Now $\{X=m, \text{final roll is }a\}$ happens iff the final roll is $a$, and among the first $m-1$ rolls, there are exactly $9 \ a$, there are $< 10\ b$ and $<10 \ c$.  And of course $P(\text{final roll is }a) = 1/3$ and it is independent of what happened before, so:


*

*$P(X=m, \text{final roll is } a) = {1\over 3} P(\text{exactly $9\ a$ and $<10 \ b$ and $<10 \ c$ among first $m-1$ rolls})$
Combining, we have:


*

*$P(X=m) = P(\text{exactly $9\ a$ and $<10 \ b$ and $<10 \ c$ among first $m-1$ rolls})$
For $m \le 19$, after having exactly $9 \ a$ among the first $m-1$ rolls, the number of non-$a$ rolls left $= m-1-9 = m-10 \le 9$, so the requirements on $b,c$ are automatically satisfied.  So:
$$\text{For } m \le 19: P(X=m) = {m-1 \choose 9} 2^{m-10} / 3^{m-1}$$
as in the answer by @ploosu2
For $m \in [20, 28]$ we start having the possibility of $b,c \ge 10$ which we must rule out.  I don't know of a way to rule them out without using another summation, but the number of cases is linear (not e.g. quadratic).  Specifically, there are $m-1-9 = m-10$ non-$a$ rolls and let $y = $ no. of $b$, so $m-10-y =$ no. of $c$.  Then the valid cases require $y \le 9$ and $m-10-y \le 9$ (equiv.: $y \ge m-19$), which is a continuous range:
$$\text{For } m \in [20, 28]: P(X = m) = {m-1 \choose 9} \sum_{y=m-19}^9 {m-10 \choose y} / 3^{m-1}$$
Or if you prefer to count the excluded cases then those are $y \ge 10$ or $m-10-y \ge 10$ (equiv.: $y \le m-20$).  Writing the restricted cases (informally) as $R = (\{y \ge 10\} \cup \{y \le m-20\}) \cap \{0 \le y \le m-10\}$ we have:
$$\text{For } m \in [20,28]: P(X = m) = {m-1 \choose 9} \left(2^{m-10} - \sum_{y \in R} {m-10 \choose y}\right) / 3^{m-1}$$
Just to verify: for the special case of $m=20$ we have $R = \{0, 10\}$ and: 
$$P(X=20) = {{19 \choose 9} (2^{10} - {10 \choose 0} - {10 \choose 10}) \over 3^{m-1}} = {{19 \choose 9}(2^{10} - 2) \over 3^{m-1}} = {{19 \choose 9}2^{10} - {20 \choose 10} \over 3^{m-1}}$$
because ${20 \choose 10} = {20 \over 10} {19 \choose 9} = 2 {19 \choose 9}$, and this also matches the answer by @ploosu2
