Probability of absorption of a biased random walk when the starting point has binomial distribution 
Consider a random walk $\{0,1, ... , N\}$ with up probability $p$ and down probability of $p-1$ where $p \neq 1/2$. Suppose there are absorbing barriers at $0$ and $N$ and that the starting point, $S_0$, is random and has a $Bin (N,\theta)$ distribution for some $\theta \in (0,1)$. By conditioning on the starting point, find the probability of absorption at $N$.

The question seems to be asking to find 
$$P(\text{absorption at}\ N|S_0=k)=P(\text{absorption at}\ N, S_0=k)/P(S_0=k)$$
The numerator is the normal probability of absorption at $N$. Thus,
$$P(\text{absorption at N}\ |S_0=k)=\frac{\frac{1-(\frac{1-p}{p})^k}{1-(\frac{1-p}{p})^N}}{\binom{N} {k} \theta^k(1-\theta)^{n-k}}$$
This, however, does not seem like the right direction. Any suggestions?

EDIT What occurs to me is that this likely asking for a probability generating function. However, I am not sure how to implement it.
 A: I shall try to give you a short answer (in yellow boxes) as well as a commentary on on how to get it. The first step is just to find the probability of absorption given a starting point. We do this with recurrence relationships.

Let $a_n$ be the probability of being absorbed at $N$ given that the starting point is at $n$.

The reason I did this is because this is the way I would normally approach a problem about absorption probabilities and, knowing how the solution usually goes, I think I can work in the binomial bit at the end.

We define $a_n$ only for $n=0,1,\dots,n$. We get the recurrence relation
  $$a_n=pa_{n+1} + (1-p)a_{n-1}$$
  From the statement of the problem. We also get the boundary conditions $a_0=0,\,a_N=1.$

The recurrence relationship comes from the fact that once you've moved up (or down) one unit the problem looks just the same as at the beginning but with a different starting point.

Therefore $(1-p)a_n - a_{n+1} + pa_{n+2}=0.$ We solve this relation by standard methods:
  Let $a_n=\lambda^n.$ Then $a_{n+1} = \lambda a_n,\; a_{n+2}=\lambda^2 a_n.$ Substituting into the recurrence relation, we get 
  $$(1-p) - \lambda + p\lambda^2=0$$ which can be factorised into
  $$\left(\lambda -1\middle)\middle(p\lambda - \left(1-p\right)\right)=0$$
  so the general solution is:
  $$a_n=A+B\left(\frac1p-1\right)^n$$
  for arbitrary constants $A,B$. Using our boundary conditions,
  $$0=A+B,\qquad 1=A +B\left(\frac1p-1\right)^N$$
  Therefore,
  $$A=-B,\qquad B=\frac1{\left(\frac1p-1\right)^N-1}$$
  So
  $$\mathbb P(\text{absorption at N | started at n}) = a_n = \frac{\left(\frac1p-1\right)^n-1}{\left(\frac1p-1\right)^N-1},$$
  except in the case that $p=\frac12$ which we will cover later.

Trying some approximate values for $p$ gives reasonable answers. We can’t solve like this for $p=\frac12$ because the recurrence relation is degenerate.
We now plug in the binomial probability for the starting point:

Therefore we get \begin{align}\mathbb P(\text{absorption at N}) &=\sum_{n=1}^N\binom Nna_n\theta^n(1-\theta)^{N-n}\\
&=\frac1{\left(\frac1p\right)^N-1}\left(\sum\binom Nn\left(\frac1p-1\right)^n\theta^n(1-\theta)^{N-n} - \sum\binom Nn\theta^n(1-\theta)^{N-n}\right)
\\&=\frac1{\left(\frac1p\right)^N-1}\left(\left(\left(\frac1p-2\right)\theta + 1\right)^N - 1\right).\end{align}
Now consider $p=\frac12.$ We have $a_n=A+B n$ so with the boundary conditions:
  $$a_n=\frac nN$$
  If $G(x)$ is the pgf for the distribution for our starting point, we get
  $$\mathbb P(\text{absorption at N})=\frac{G'(1)}N.$$
  As we know $G(x)=\left((1-\theta) + \theta x\right)^N,$ we get the surprising answer of:
  $$\mathbb P(\text{absorption at N})=\theta$$

