Urns version of Laplace's law of succession I'm trying to prove urns version of Laplace's law of succession my professor suggested. Laplace's law states that the chance that the next trial is a success given $j$ successes out of the first $n$ is $\frac{(j+1)}{(n+2)}$. Here is how the problem states: 
"If we have $n+k+1$ urns and urn $i$ has $i$ balls labeled $1$ and $n+k-i$ labeled zero. We pick an urn at random and draw $n$ balls from it without replacement say $j$ of them are ones. Show that the conclusion of Laplace's law holds for this setup. In other word, the chance that the next ball is a one is $\frac{(j+1)}{(n+2)}$."
I've proved one version of the law, which close to this version An elementary version of Laplace's Method of Succession. I tried to use similar approach but somehow my answer always in form of $k$ and I can't get rid of it. What is the intuition behind this $k$? Is it just to increase the complexity of the problem, or it has some meaning behind it?     
 A: Let $I$ be the random variable indicating the label of the urn picked, $J$ the number of ones picked (after picking $n$ balls), and $X$ the event that the next ball is one. We know that
$$
P(J = j \mid I = i) = \frac{\binom{i}{j}\binom{n+k-i}{n-j}}{\binom{n + k}{n}}
= \frac{\binom{j+(i-j)}{i-j}\binom{n+k-j-(i-j)}{k-(i-j)}}{\binom{n+k}{n}}
$$
and
$$
P(X \mid I = i, J = j) =
\begin{cases}
0 & ; i \le j \\
\frac{i-j}k & ; j \le i \le k + j
\end{cases}
$$
Compute $P(I = i \mid J = j)$ from
$$
P(I = i \mid J = j) = \frac{P(I = i, J = j)}{\sum_{\tilde i}P(I = \tilde i, J = j)} =
\frac{P(J = j \mid I = i)}{\sum_{\tilde i} P(J = j \mid I = \tilde i)}
$$
where we use $P(I = i) = \frac{1}{n + k + 1}$ in the last equation. Then,
\begin{align*}
P(X \mid J = j) & = \sum_{i} P(X \mid I = i, J = j) P(I = i \mid J = j) \\
& = \sum_{i=j}^{k+j} \frac{i-j}k \frac{P(J = j \mid I = i)}{\sum_{\tilde i} P(J = j \mid I = \tilde i)} \\
& = \frac{\sum_{i=j}^{k+j}(i-j)P(J = j \mid I = i)}{k \sum_{i=j}^{k+j} P(J = j \mid I = i)} \\
& = \frac{\sum_{i=j}^{k+j}(i-j)\binom{i}{j}\binom{n+k-i}{n-j}}{k \sum_{i=j}^{k+j} \binom{i}{j}\binom{n+k-i}{n-j}} \\
& = \frac{\sum_{i=0}^k i \binom{j+i}{i}\binom{n-j+(k-i)}{k-i}}{k\sum_{i=0}^k \binom{j+i}{i}\binom{n-j+(k-i)}{k-i}}
\end{align*}
I don't know how to make this into the desired expression $\frac{j+1}{n+2}$. Putting $k = 1$ seems to work though.
