Suppose that we have started with $w$ white balls and $b$ black balls. Then
\begin{align*}
\mathbb{P}(w_n = w+k)
&= \binom{n}{k} \frac{w(w+1)\dots(w+k-1)b(b+1)\dots(b+n-k-1)}{(w+b)(w+b+1)\dots(w+b+n-1)} \\
&= \frac{1}{B(w, b)} \binom{n}{k} \frac{\Gamma(w+k)\Gamma(b+n-k)}{\Gamma(w+b+n)} \\
&= \frac{1}{B(w, b)} \frac{k^{w-1} (n-k)^{b-1}}{n^{w+b-1}} \frac{E_k(w)E_{n-k}(b)}{E_n(b+w)} ,
\end{align*}
where $\Gamma(\cdot)$ is the gamma function, $B(\alpha, \beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$ is the beta function, and
$$ E_n(z) := \frac{\Gamma(n+z)}{n!n^{z-1}}. $$
Note that $E_n(z) \to 1$ as $n\to\infty$. So, if we write $p_k = k/n$, then the m.g.f. of $X_n$ is explicitly given by
\begin{align*}
\mathbb{E}[e^{\lambda X_n}]
= \frac{1}{B(w, b)} \sum_{k=0}^{n} \exp\biggl( \lambda \frac{p_k + w/n}{1 + (w+b)/n} \biggr) p_k^{w-1}(1 - p_k)^{b-1} \frac{1}{n} \cdot \frac{E_k(w)E_{n-k}(b)}{E_n(b+w)}.
\end{align*}
Letting $n \to \infty$, this converges to
\begin{align*}
\mathbb{E}[e^{\lambda X_{\infty}}]
= \frac{1}{B(w, b)} \int_{0}^{1} e^{\lambda p} p^{w-1}(1 - p)^{b-1} \, \mathrm{d}p.
\end{align*}
From this, we read out that the distribution of $X_{\infty}$ has the density
$$ f(p) = \frac{1}{B(w, b)} p^{w-1}(1 - p)^{b-1} \mathbf{1}_{(0,1)}(p), $$
proving that the limit distribution is $\operatorname{Beta}(w, b)$.
Best Answer
I haven't gone through the math for $\mathbb{P}(W_n) = \mathbb{P}(W_1)$, but assuming it is correct, we have
$$\mathbb{P}(W_1 | W_n) = \mathbb{P}(W_n | W_1)\cdot\frac{\mathbb{P}(W_1)}{\mathbb{P}(W_n)} = \mathbb{P}(W_n | W_1)$$
Now let's use our common sense. Given the event $W_1$ happened, that means that we now have $w + 1$ whites and $b$ blues after the first extraction. Then, we have $(n - 1)$ more extractions to go. Therefore, if we extend our notation of $W_n$ a bit to $W_{n,w,b}$ naturally, then
$$\mathbb{P}(W_n | W_1) = W_{n - 1, w + 1, b} = W_{1, w + 1, b} = \frac{w + 1}{w + b + 1}$$
At least, I believe.
P.S. I am currently
procrastinatingrevising for my probability exam next week, and this is precisely what I am studying! :)