Odds of Bankruptcy Flipping a Weighted Coin Infinity Times A random tangent I have been thinking about, please be gentle I am not a mathematician:
Imagine a coin toss where Heads means we win a coin and Tails means we lose a coin.  We start with 1 coin and cannot go negative (i.e. if we flip a Tails on the first try, we lose and it's game over).
The coin is weighted so that it flips Tails with a probability of a.  If a = 0, the probability of bankruptcy is 0%.  At some value between 0 and 1, the probability of bankruptcy approaches 100%.  For what value of a does this probability cross 50%?
So I drew a graph of the outcomes for three flips of a coin with a = 0.4:
       1
(0.4) / \ (0.6)
     0   2
  (0.4) / \ (0.6)
       1   3
      / \ / \
     0   2   4

First Flip: P(0) = 40%, P(2) = 60%
Second Flip: P(0) = 40%, P(1) = 24%, P(3) = 36%
Third Flip: P(0) = 49.6%, P(2) = 28.8%, P(4) = 21.6%

The first thing I noticed is the probability of bankruptcy on evened-numbered flips (e.g. flip #2, flip #4, ...) does not increase.  So you have a 40% of busting on the first hand, the probability remains unchanged on the second flip.
I attempted to try and solve this problem in a closed form solution shown below (sorry it's a picture, I don't know LaTeX...).

For a = 1/2, I got a P(0) = 2/3.  Obviously this is wrong because for a equally weighted coin, we would expect that playing infinitely long, you would succumb to gambler's ruin pretty quick.
So again restating the original question: How weighted does the coin have to be in our favor that after playing for an arbitrarily long amount of time, our odds of bankruptcy approaches 50%?  Is this a problem that can be solved in closed form?  Also if you can point out the fallacies in my attempt, I greatly appreciate it.
Thanks in advance!
 A: Here's a combinatorics way of doing the problem. Suppose $p$ is the probability of losing on any given flip.
Then the total probability of ruin is given by
$$\sum_{n=0}^{\infty}C_np^{n+1}(1-p)^n$$
where $C_n$ counts the number of sequences of wins and losses on flips such that you lose your last coin on exactly turn $2n+1$.
This sort of restricted random walk shows up everywhere in combinatorics and the count $C_n$ is known as the Catalan numbers, with formulas such as $C_n=\frac{1}{2n+1}\binom{2n}{n}=\binom{2n}{n}-\binom{2n}{n+1}$.
In particular (see this link for a proof), we know the generating function for the Catalan numbers, that is to say
$$g_C(x)=\sum_{n=0}^{\infty}C_nx^n=\frac{1-\sqrt{1-4x}}{2x}\text{.}$$
Now we can just simplify the probability of ruin as
$$\sum_{n=0}^{\infty}C_np^{n+1}(1-p)^n=pg_C\bigl(p(1-p)\bigr)=p\frac{1-\sqrt{1-4p(1-p)}}{2p(1-p)}=\frac{1-\left|2p-1\right|}{2(1-p)}=\begin{cases}1&p\geq\frac{1}{2}\\\frac{p}{1-p}&p\leq\frac{1}{2}\end{cases}\text{,}$$
where gambler's ruin manifests in the absolute value.
If specifically solving for $\Pr($ruin$)=\frac{1}{2}$, $p=\frac{1}{3}$.
