would definitely lose all after gambling for a long time?

Question

I have a mathematic question about gambling.

Given that Peter is gambling and

1. the winning probability $50\%$.
2. gain \$1 for winning, lose \$1 for loss
3. Initial capital: $\$A$
4. the game will end either capital reach $\$B$ or Peter lost all.
5. $n$ is the number of game playing.
6. $P(n)= \text{[probability of lost all]}$

Therefore,

$$ P(0)=1; \qquad P(B)=0 $$

and

$$ \begin{array}{rc} & P(n) = 50\% \cdot P(n-1) + 50\% \cdot P (n+1) \\ \implies \quad & 2P(n) = P(n-1) + P(n+1) \\ \implies \quad & P(n)-P(n-1) = P(n+1) - P(n). \end{array} $$

$ ``\text{change of $P$''} = 1/B $

Therefore $P(A) = 1-A/B$.

If $B = \infty$, then $P(\infty) = 1$.

My question is, how about if the game winning probability is $60\%/40\%$ instead of $50\%/50\%$? How should I calculate?

Thank you a lot.