Let $X$ be the random variable denoting the total number of drawn marbles. We are interested in the expected number of drawn marbles, i.e. $\mathbb{E}[X]$. If we label our colours with the numbers $1,..,k$ then $X=\sum_{j=1}^kX_j$ for $X_j$ being the number of drawn colour-$j$ marbles. Note that $X_j$ is distributed according to a multivariate hypergeometric random variable, where the total number of draws is also a random variable $X$ and to be determined. Hence
\begin{eqnarray}
\mathbb{E}[X] &=& \mathbb{E}[\sum_{m=n}^{k(n-1)+1}m\chi_{X=m}] \nonumber \\
&=& \sum_{m=n}^{k(n-1)+1}m\mathbb{P}[X=m] \nonumber \\
&=&\sum_{m=n}^{k(n-1)+1}m\mathbb{P}[\exists!X_j=n\text{ and last drawn marble is of colour }j]. \nonumber
\end{eqnarray}
Now all left to do is to figure out the number $A_m=\mathbb{P}[\exists!X_j=n\text{ and last drawn marble is of colour }j]$. We have that $X_j$ is a multivariate hypergeometric random variable with $m$ draws, $k$ colours all of which we have $n$ marbles:
This means that the probability of drawing $(a_j)_{j=1,..,k}$ marbles of each colour $j$ with a total of $m$-marbles is precisely
$$
\mathbb{P}_{\text{Hypergeometric}_{m,n,k}}[\text{Draw } (a_j)_{j=1,..,k}]=\frac{\binom{n}{a_1} ... \binom{n}{a_k}}{\binom{nk}{m}}
$$
A first guess would be
\begin{eqnarray}
A_m &"="& k \sum_{\substack{a_1,..,a_{k-1}=0 \\ a_1+...a_{k-1}+n=m}}^{n-1}\frac{\binom{n}{a_1}...\binom{n}{a_{k-1}}}{\binom{nk}{m}} \nonumber. \\
\end{eqnarray}
However this does not rule out the cases, where we have not drawn colour $j$ lastly. To account for this we simply draw one marble less of colour $j$ and sample the remaining.
\begin{eqnarray}
A_m &=& k \frac{1}{nk-m+1} \sum_{\substack{a_1,..,a_{k-1}=0 \\ a_1+...a_{k-1}+n=m}}^{n-1}\frac{\binom{n}{n-1}\binom{n}{a_1}...\binom{n}{a_{k-1}}}{\binom{nk}{m-1}} \nonumber\\
&=& \sum_{\substack{a_1,..,a_{k-1}=0 \\ a_1+...a_{k-1}+n=m}}^{n-1}\frac{\binom{n}{a_1}...\binom{n}{a_{k-1}}}{\binom{nk-1}{m-1}}.\nonumber\\
\end{eqnarray}
Thus we have the explicit formula
$$
\mathbb{E}[X]=\sum_{m=n}^{k(n-1)+1}mA_m.
$$
Edit 1: Using Vandermonde's identity (for binomial coefficients) we (almost) have that
\begin{eqnarray}
A_m &"="& \frac{\binom{n(k-1)}{m-n}}{\binom{nk-1}{m-1}} .\nonumber\\
\end{eqnarray}
However note that the indices $a_j$ cannot attain the value $n$.
Best Answer
Assume we draw from the bag until the bag is empty. What's the probability that blue is last to be exhausted? We denote this event as $B_3$, ie. blue is the third color to be exhausted from the bag. Applying the same logic as the two color case, we get: $$P(B_{3}) = \frac{b}{r+g+b}$$
Let's assume we already know we are in a situation where blue is the longest survivor. What then would be the probability that green is the second to be exhausted? The second longest-survivor is either green or red, so for all intents and purposes, we can ignore blue. We get the conditional probability:
$$P(G_{2}|B_{3}) = \frac{g}{r+g}$$
From these we can get probability that "green is second to be exhausted and blue is third":
$$P(G_{2} \land B_{3}) =P(B_{3}) P(G_{2}|B_{3}) = \\ = \left(\frac{b}{r+g+b} \right) \left(\frac{g}{r+g} \right)$$
We may compute:
$$P(R_{1}) = P(G_{2} \land B_{3}) + P(B_{2} \land G_{3}) = \\ = \left(\frac{b}{r+g+b} \right) \left(\frac{g}{r+g} \right) + \left(\frac{g}{r+g+b}\right) \left(\frac{b}{r+b} \right) $$
Applying to $r = 1$, $g = 2$, $b = 3$, we get: $$ P(R_1) = \frac{2}{1+2+3}\;\frac{3}{1+3} + \frac{3}{1+2+3}\;\frac{2}{1+2} = \frac{7}{12} $$