= Solution
Let $B$ count the red balls among the $n$ added balls. By <independence of random variables>, $B$ has a <binomial distribution> with parameters $n,1/2$. The total red count is $R=1+B$, and the box has $n+1$ balls. Conditional on $R=r$, <sampling without replacement> gives
$$
\mathbb P(D_1\mid R=r)=\frac r{n+1},\qquad\mathbb P(D_1\cap D_2\mid R=r)=\frac{r(r-1)}{n(n+1)},
$$
where $D_1,D_2$ denote red first and second draws. Averaging these <conditional probabilities> and taking their ratio gives
$$
\mathbb P(D_2\mid D_1)=\frac{\mathbb E[R(R-1)]}{n\mathbb E[R]}.
$$
The <binomial distribution> moments give $\mathbb E[B]=n/2$ and $\mathbb E[B^2]=n/4+n^2/4$. Consequently
$$
\mathbb E[R]=\frac{n+2}{2},\qquad\mathbb E[R(R-1)]=\mathbb E[B^2+B]=\frac{n(n+3)}4.
$$
Therefore the <red-ball prediction with one certain red ball> is
$$
\boxed{\mathbb P(D_2\mid D_1)=\frac{n+3}{2(n+2)},\qquad n\geq1.}
$$
It reduces to $2/3$ at $n=1$ and approaches $1/2$ as the number of randomly coloured extra balls grows. Taking an unweighted average of the remaining red fractions would miss the change in weights caused by taking a <conditional probability> given the first red draw.
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