Random Walk: Gambler's Ruin - Infinite Fortune Given that one starts at $i$ dollars, how does one find the probability of getting infinitely rich/getting ruined (the complement)?
Let's say that $p = \text{probability of winning an extra dollar}$, and $q = \text{probability of losing a dollar}$. The gambler is ruined once he has $0$ dollars (I think the assumption is that the opposing party has unlimited dollars).
I found the formula online, which is that if $p \leq \frac{1}{2}$, then the probability of winning becomes $0$. However, if $p \geq \frac{1}{2}$, then the probability of winning is not $1$, but follows this formula:
$P(\text{becoming infinitely rich, given that you start with $i$ dollars}) = 1 - (\frac{q}{p})^i$.
What's the intuition behind this?
 A: Following Herb Steinberg's comment, I get a different value than the one in the question, so I'm posting my calculation.
As Herb has pointed out, the probability of going broke is $$q^i\sum_{n=0}^\infty\binom{2n+i}{n}(pq)^n$$ since the gambler will have to lose $i$ more games than he wins in order to lose all his money.  This suggests that we compute the value of $$P_i(x)=\sum_{n=0}^\infty\binom{2n}nx^n$$
It is shown here that $$P_i(x)=
\frac {1}{\sqrt {1 - 4x}} \left( \frac {1 - \sqrt {1 - 4x}}{2x} \right)^i,$$ but I'll include a slightly adapted version of the proof, for completeness.  We have $$\begin{align}
P_i(x)&=\sum_{n=0}^\infty\binom{2n}nx^n\\
&=\sum_{n=0}^\infty\left[\binom{2n+i-1}n+\binom{2n+i-1}{n-1}\right]x^n\\
&=\sum_{n=0}^\infty\binom{2n+i-1}nx^n+\sum_{n=0}^\infty\binom{2n+i-1}{n-1}x^n\\
&=P_{i-1}(x)+\sum_{n=0}^\infty\binom{2(n-1)+(i+1)}{n-1}x^n\\
&=P_{i-1}(x)+xP_{i+1}(x)
\end{align}$$
where we have used the usual convention $\binom{n}{-1}=0$.
Now $$P_0(x)=\sum_{n=0}^\infty\binom{2n}{n}x^n=\frac1{\sqrt{1-4x}},\ |x|<\frac14$$ which is easily seen by applying the binomial  formual to $(1-4x)^{-1/2}$.
Now we specialize to $x=pq$ and we drop the dependence on $x$, so that $$P_0=\frac1{\sqrt{1-4pq}}=\frac1{\sqrt{4p^2-4p+1}}=\frac1{2p-1},$$ since $p>\frac12$.
Also, $$\begin{align}
P_1&=\sum_{n=0}^\infty\binom{2n+1}{n}(pq)^{n}\\
&=\frac12\sum_{n=0}^\infty\binom{2n+2}{n+1}(pq)^n\\
&=\frac12\sum_{n=1}^\infty\binom{2n}{n}(pq)^{n-1}\\
&=\frac12\left(\frac{P_0-1}{pq}\right)\\
&=\frac1{2pq}\left(\frac1{2p-1}-1\right)\\
&=\frac1{2pq}\left(\frac{2-2p}{2p-1}\right)\\
&=\frac1{2p-1}\frac1p
\end{align}$$
Now it's easy to verify that $$P_i=\frac1{2p-1}\frac1{p^i}$$ satisfies the recurrence $$P_i=P_{i-1}+pqP_{i+1}$$ and since the initial values $P_0,P_1$ also satisfy this formula the recurrence is solved.  Going back to the original formula for the probability of going broke, starting with a bankroll of $i$ is $$
\frac1{2p-1}\left(\frac qp\right)^i,$$ and in contrast to the formula in the question, the probability of never going broke is $$\boxed{1-\frac1{2p-1}\left(\frac qp\right)^i}$$
