Your question can be described via a simple random walk on $\mathbb Z$: Choose i.i.d. random variables $X_1, X_2, \ldots$ such that
$$P[X_i = +1] = P[X_i = -1] = \frac 12$$
$X_i$ describes the outcome of the $i$-th toss of a coin: $+1$ means we win one dollar, $-1$ we lose one dollar. Furthermore, let $S_k = \sum_{i=1}^k X_i$ be the amount of money lost or won until time $k$.
Now, for a positive integer $n$ (which describes the initial amount of money we own), define $$T_{-n} = \inf\{k\in \mathbb N\mid S_k = -n\}$$ to be the number of coin tosses until we go bankrupt. Then we have
Lemma: $P[T_{-n} > k] = \Theta(1/\sqrt k) \qquad \text{ as }k\to \infty$
which is to say: The probability of not being bankrupt after the $k$-th toss of a coin decreases like $1/\sqrt{k}$ (in particular, this goes to zero; so we will go bankrupt, eventually). On the other hand the calculation $$E[T_{-n}]= \sum_{k=1}^\infty P[T_{-n}>k] \approx \sum_{k=1}^\infty \frac{1}{\sqrt{k}} = \infty$$ shows
Corollary: $E[T_{-n}] = \infty$
which says, we may well have to wait a very very long time before we go bankrupt.
In an attempt to be self-contained. Here is
Proof of the Lemma: The proof involves two steps:
Claim 1: We have $$P[S_n = 2k-n] = \binom{n}{k}2^{-n}$$ for $k\le n$, and $P[S_n = x] = 0$ for all other $x$.
and
Claim 2: For $k>0$ we have $$P[T_{-n} \le k] = P[S_k \notin (-n, n]\,] $$
For Claim 1 just note that in order for $S_n = 2k-n$, we need to win $k$ times and lose $n-k$ times and there are $\binom nk$ possibilites to choose $k$ winners out of $n$.
For Claim 2 write $P[T_{-n} \le k] = \sum_{b = -\infty}^\infty P[T_{-n} \le k, S_k = b]$ and notice that for $b > -n$ we have $$P[T_{-n} \le k, S_k = b] = P[S_k = -2n - b]$$ The last assertion is obtained by reflecting each path visiting $-n$ and ending at $b$ at the first time it hits $-n$. So that for the second part of its path (after first hitting $-n$), all the values of the $X_i = \pm 1$ get swapped to $X_i = \mp 1$. This gives a one-to-one correspondence between paths to $b>-n$, which visit $-n$ and paths to $-2n-b$. (I'm sure this explanation for Claim 2 is hardly understandable, but I can't come up with a better explanation... This is sometimes called the reflection principle). So
\begin{align}
P[T_{-n} \le k] &= \sum_{b = -\infty}^\infty P[T_{-n} \le k, S_k = b] \\
&= \sum_{b\le -n} P[S_k = b] + \sum_{b> -n} P[S_k = -2n -b] \\
&= P[S_k = -n] + 2P[S_k < -n] \\[6pt]
&= P[S_k \notin (-n,n]\, ]
\end{align}
Therefore (even values of $k$ suffice by monotonicity)
\begin{align}
P[T_{-n} > 2k] &\ge P[S_{2k} = 0\, ] = \binom{2k}{k} 2^{-2k} \sim \frac{1}{\sqrt{\pi k}} \\
P[T_{-n} > 2k] &\le 2n \cdot P[S_{2k} = 0\, ] = 2n\cdot \binom{2k}{k} 2^{-2k} \sim \frac{2n}{\sqrt{\pi k}}
\end{align}
q.e.d.
