Consider the following well-known betting strategy, sometimes called the Martingale strategy: A gambler bets one unit. If he wins, he again bets one unit. If he loses, he bets double the amount that he just lost so that, when he eventually wins, he is again ahead one unit.
Suppose the gambler continues betting forever, or until he is ruined, whichever comes first. Is there a closed-form expression for the probability that the gambler is ruined, given that he starts with some initial amount n and wins each game independently with probability p > .5?
If there is no closed-form expression for P(ruin), I will settle for finding the smallest value for p for which P(ruin) < 1. Clearly P(ruin) = 1 when p = .5 and P(ruin) = 0 when p = 1, but it's not clear to me what happens in between.
For convenience [edit: and for other reasons; see Addendum below] I was assuming that the gambler is ruined when he is forced to abandon his betting strategy regardless of whether he has money remaining and that $n = 2^k - 1$ so that he is ruined if he loses k consecutive bets before reaching $n=2^{k+1}-1$; but you may assume any reasonable definition of ruin and initial value for n that you like (i.e., whatever makes the math easier).
Under the above assumptions, this is what I have so far: Consider a series of k games. The gambler is ruined if he loses all k games, which happens with probability $(1-p)^k$. Otherwise, he wins 1 unit with probability $1-(1-p)^k$. He must win $2^k$ consecutive series to reach the next betting level, which happens with probability $[1-(1-p)^k]^{2^k}$. In order to avoid eventual ruin, he must avoid ruin at each betting level, which happens with probability $\prod_{j=k}^{\infty}[1-(1-p)^j]^{2^j}$, so that the overall chance of ruin is
$$\Bbb P(\text{"ruin"}) = 1 - \prod_{j=k}^{\infty}[1-(1-p)^j]^{2^j}$$
...but I'm not having any luck getting the above into a closed-form expression. As mentioned above, I'm not even sure for which values of p the probability is less than 1.
Edit/Update: I don't really need the answer to this problem any longer, though I'll leave it here since I am still curious if someone has a more complete solution. I was not able to find a closed-form expression, but I was able to show convergence as follows. Using $q = 1 - p$ and $r=\lim_{n \to \infty}\prod_{j=k}^{n}(1-q^j)^{2^j}$ for convenience:
$$ln(r)=\lim_{n \to \infty}\sum_{j=k}^n 2^j ln(1-q^j) \\ =\lim_{n \to \infty}-\sum_{j=k}^n 2^j \sum_{i=1}^\infty \frac{(q^j)^i}{i} \\ =\lim_{n \to \infty}-\sum_{j=k}^n \sum_{i=1}^\infty \frac{(2q^i)^j}{i} \\ =-\sum_{i=1}^\infty \sum_{j=k}^\infty \frac{(2q^i)^j}{i} \\ =-\sum_{i=1}^\infty \frac{\frac{(2q^i)^k}{1-2q^i}}{i} \\ =-\sum_{i=1}^\infty2^k\frac{(q^k)^i}{i(1-2q^i)}$$ and the last sum converges to some (positive) value s since q < .5. Thus $ln(r) = -s$, $r = e^{-s}$, and $\Bbb P(\text{"ruin"}) = 1 - e^{-s}$ for some positive value s. This shows that the chance of ruin is strictly less than 1 for all values p > .5. The gambler might continue betting forever.
Addendum: One might wonder why I have chosen a somewhat nontraditional definition of ruin, choosing "ruin" to refer to the point at which the gambler must abandon his strategy rather than the point at which he is absolutely bankrupt. By selecting such a definition, however, we make our result much more general: the calculation above provides an upper bound for all possible follow-up strategies the gambler may adopt, no matter how esoteric. Furthermore, since it is trivially easy to select a follow-up strategy that ruins the gambler almost surely (betting his entire fortune at every step will suffice), the calculation above proves that this upper bound is in fact the least upper bound. Thus no matter what follow-up strategy the gambler may adopt after being forced to abandon his initial strategy, the probability of ruin is never greater than $1-e^{-s}$.
The only thing remaining is to determine, if possible, a closed-form expression for $s=2^k\sum_{i=1}^\infty\frac{(q^k)^i}{i(1-2q^i)}$.