When will a community end? Let us imagine a hypothetical community, which consists of a large number of households, say $N=1000$.
 Every household has an initial capital, say $a=10$ capital units so that the combined capital of the community is $Na=10000$ capital units.
 At the end of the every year every household wins or loses a capital unit with probabilities $\frac{1}{2}$ and $\frac{1}{2}$ respectively.
If a household's capital is reduced to zero the household is ruined.  
We are interested in the expectation of the time (in years) of the ruin of the whole community. (The time of the ruin of the last remaining household.)  
Edit: 
Did is right. The expectation is an infinity. But let us make things more interesting. Suppose that the probability of the win is $p=\frac{1}{3}$ and the probability of the lose is $q=\frac{2}{3}$.
 A: Each household $h$ is ruined after a random time $T_h$ which is almost surely finite but with infinite expectation. The whole community is ruined after a random time $T$. Choose any household $h$. Then $T\geqslant T_h$ hence $\mathbb E(T)$ is infinite.
To show that each $T_h$ has infinite expectation, recall that the time $\tau(a,b)$ to reach $0$ or $b\gt a$ for a simple symmetric random walk starting from $a$ is such that $\mathbb E(\tau_{a,b})=a(b-a)$. And $\mathbb E(T_h)\geqslant\mathbb E(\tau(a,b))$ for every $b\gt a$.
Edit: When loss is favored, each household $h$ is ruined after a random time $T_h$ which is the sum of $a$ i.i.d. copies of the time $\tau$ when its capital is its initial capital minus $1$. If the probability to win is $p\lt\frac12$, after one step, either $\tau$ is reached, or one needs two copies of $\tau$ to reach one's initial capital minus $1$. Hence, for every $|s|\leqslant1$, $u(s)=\mathbb E(s^\tau)$ solves $u(s)=s(1-p+pu(s)^2)$, hence
$$
u(s)=\frac{1-\sqrt{1-4p(1-p)s^2}}{2ps}.
$$
The whole community $C$ is ruined at time $T=\max\{T_h\mid h\in C\}$ hence, for every $k\geqslant1$,  $\mathbb P(T\leqslant k)=\mathbb P(T_h\leqslant k)^N$ where the distribution of $T_h$ is characterized by the identity 
$$
\mathbb E(s^{T_h})=u(s)^a.
$$
Furthermore,
$$
\mathbb E(T)=\sum_{k\geqslant1}\mathbb P(T\geqslant k)=\sum_{k\geqslant1}\left(1-\mathbb P(T_h\leqslant k)^N\right).
$$
It seems difficult to deduce the exact value od $\mathbb E(T)$ from these formulas. Empirically, one can probably estimate $\mathbb E(T)$ by $k$ where $\mathbb P(T_h\geqslant k)\approx1/N$. Now, $\tau$ is exponentially integrable hence, when $k$ is large, $\mathbb P(\tau\geqslant k)\approx r^k$ where $r=\sqrt{4p(1-p)}$. Since $T_h$ is the sum of $a$ copies of $\tau$, one can probably estimate $\mathbb P(T_h\geqslant k)$ by  $k^{a-1}r^k/(a-1)!$.
For $p=\frac13$, $a=10$ and $N=1000$, this would yield $\mathbb E(T)\approx950$.
A: It seems that as a first approximation you can reduce the analysis to a single household, asking how many years until it is ruined with probability $(1/2)^{1/N}$, or roughly 99.93% for N = 1000.
