TITLE: Chance of losing a biased "Random Walk" game
QUESTION [5 upvotes]: Consider a game where you start with 1 point.
Then you flip a fair coin infinitely many times.
For each heads, you gain 2 points.
For each tails, you lose 1 point.
What is the probability that your score never goes below 1?
edit: I am looking specifically for the answer with flipping infinitely many times, not the probability after $N$ flips as $N$ approaches infinity.

REPLY [4 votes]: One way to solve this is to let $p_n$ be the probability that, if you start with $n$ points, you eventually zero points. Observe that $p_1$ is also the probability that if you start with $n$ points, you ever have $n-1$ points.
Using this, one can see that $p_n=p_1p_{n-1}$, since the probability of getting from $n$ to zero equals the probability of eventually getting to $n-1$, then the probability of getting from there to $0$. In particular, we get that $p_n=p_1^n$ by using this argument.
Then, observe that $p_n=\frac{p_{n+2}+p_{n-1}}{2}$ by looking at what happens after the first step. Substituting in at $n=1$ gives $p_1=\frac{p_1^3+1}2$ gives that $p_1$ has to be either $1$ or $\frac{-1\pm\sqrt{5}}{2}$. Clearly $p_1$ cannot be $\frac{-1-\sqrt{5}}2$ because that's negative. Thus $p_1$ is either $1$ or $\frac{-1+\sqrt{5}}2$.
To see that $p_1$ is the lesser value, we can define $p_{n,t}$ to be the probability of reaching zero in at most $t$ steps. Note that $p_{n,t+1}=\frac{p_{n-1,t}+p_{n+2,t}}2$ and $p_{n,0}=0$ for $n>0$. Induction on $t$ using this formula quickly shows that $p_{n,t}\leq \left(\frac{-1+\sqrt{5}}2\right)^n$ for all $n$ and $t$. Then, $p_n=\lim_{t\rightarrow\infty}p_{n,t}\leq \left(\frac{-1+\sqrt{5}}2\right)^n$ which implies $p_1=\frac{-1+\sqrt{5}}2$
Finally, the probability you're interested in is $1-p_1$, which equals $\frac{3-\sqrt{5}}2$.