[Note:] This is an answer to the original question. The question has been changed in the meantime, and the answer no longer answers the question in its current form. The answer to the current question is trivially that the game ends with probability $1$.
Consider the same game with a profit target of €$n$. If you roll a $6$ (why is it always a $6$?), your target effectively decreases by $1$; otherwise it increases by $1$. Thus the recurrence relation for the probability $p(n)$ of the game with target €$n$ ending is
$p(n)=\frac16p(n-1)+\frac56p(n+1)\;,$
or
$p(n+1)-\frac65p(n)+\frac15p(n-1)=0$
with characteristic equation
$\lambda^2-\frac65\lambda+\frac15=0$
with solutions $\frac15$ and $1$. Thus the general solution of the recurrence is
$p(n)=c_1+c_2\left(\frac15\right)^n\;,$
and the conditions $p(0)=1$ and $p(\infty)=0$ determine $c_1=0$ and $c_2=1$, so
$p(n)=\left(\frac15\right)^n$
and
$ p(500)=\left(\frac15\right)^{500}\approx3\cdot10^{-350}\;. $