Your approach is correct.
As for how to prove the general case, consider the following trick:
Suppose that you label the initial balls with $0$. For every $n$, when you add $k$ additional balls after the $n$-th draw, you label these new balls with the number $n$.
Let $p_0 = \frac{R}{R+B}$, and for all $n > 0$ let $p_n$ be the probability that the $n$-th ball drawn is red.
Suppose that the $i$-th ball you draw has the number $j$ on it (note that $j < i$). Then the probability that it is red is equal to $p_j$. With this, you can calculate the first few $p_n$ and form a hypothesis from there that you can prove inductively.
Let $A,B,C$ denote the events that we draw "at least 1 red", "at least 1 black", "at least 1 white" ball in 7 draws.
We're looking for $A\cap B\cap C$.
Using inclusion-exclusion we get:
$$
A\cup B\cup C=A+B+C - A\cap B - A \cap C - B\cap C + A\cap B\cap C
$$
We further have that $A\cup B\cup C = \Omega = \binom{20}7$.
Since all two element cuts still are difficult to calculate, we apply inclusion-exclusion to them as well:
$$A\cup B =A+B - A\cap B $$
We then use that $A\cup B = \binom{20}7-A^C\cap B^C $ and this event is $\binom{20}7 -120$ for $A,B$ and $\binom{20}7$ else.
Finally, we can also calculate $A,B,C$ using their complement.
Substituting in we now have:
$$
A\cup B\cup C=A+B+C - A\cap B - A \cap C - B\cap C + A\cap B\cap C
\\ \Leftrightarrow\\
A\cup B\cup C=A+B+C -A-B+A\cup B + A \cup C-A-C +B\cup C - B - C + A\cap B\cap C
\\ \Leftrightarrow\\
A\cap B\cap C = -A \cup C + A - A\cup B + A\cup B\cup C + B - B\cup C + C
\\ \Leftrightarrow\\
A\cap B\cap C = -\binom{20}7 + A -(\binom{20}7 -120) + \binom{20}7 + B - \binom{20}7 + C
\\ \Leftrightarrow\\
A\cap B\cap C = -\binom{20}7 + \binom{20}7-\binom{15}7 -(\binom{20}7 -120) + \binom{20}7 + \binom{20}7-\binom{15}7 - \binom{20}7 + \binom{20}7-\binom{10}7
\\ \Leftrightarrow\\
A\cap B\cap C = 64650
$$
And so we have that $\mathbb{P}(A\cap B\cap C) = 64650/77520$.
If we denote with $\mathcal{H}_{\mathcal{n},(\mathcal{N}_1,...,\mathcal{N}_m)}\left(\{\left(a_1,\ldots,a_m\right)\}\right)$ the hypergeometric distribution on $m$ categories with category $i$ having $\mathcal{N}_i$ elements when we draw exactly $a_i$ elements of category $i$, then we can also brute-force calculate it:
$$\begin{align}
\mathbb{P}(A\cap B\cap C) &=\sum_{i+j+k = 7\\ i,j,k\ge 1}
\mathcal{H}_{\mathcal{7},(5,5,10)}\left(\{\left(i,j,k\right)\}\right)
\\
&=\sum_{i=1}^{7-2} \sum_{j=1}^{7-i-1}
\mathcal{H}_{\mathcal{7},(5,5,10)}\left(\{\left(i,j,7-i-j\right)\}\right)
\end{align}$$
Best Answer
Theory: if $reds$ is the number of red balls you've seen and $blacks$ is the number of black balls you've seen, then the distribution is:
$Beta(reds+1, blacks+1)$
This starts out as $Beta(1,1)$ which is the uniform distribution we want. As we see blacks, it shifts toward zero; as we see reds, it shifts toward one.
Does anyone know if this is right?