If you treat this as $\pm1$ random walk where the independent probability of $+1$ steps is $p$ and of $-1$ steps is $1-p$ then it is a known result that with $p \lt \frac12$ the probability of the sum of the steps ever reaching some positive integer $n$ is $\left(\dfrac {p}{1-p}\right)^n$
Here $p =0.495$. You need to quantify _realistically_ : if you want the probability or risk of ever losing a net $n$ or more to be less than some $r$ then you need $r \lt \left(\dfrac {p}{1-p}\right)^n$ and so want $n > \dfrac{\log(r)}{\log(p) - \log(1-p)}$
For some illustrative values of the risk $r$ (the first shown is $1\%$ and the fifth $0.0001\%$) you get suggested values of $n$. You can round down if you want the risk of strictly exceeding these positions to be $r$ or less.
r n
0.01 230.25
0.001 345.38
0.0001 460.50
0.00001 575.63
0.000001 690.75