# Fair gambler's ruin tail probability

I'm looking at the following variant of the fair gambler's ruin problem: The gambler starts with 1 dollar. They repeatedly flip a fair coin. Heads, +1 dollar; Tails -1 dollar. The game stops when the gambler reaches 0 dollars.

It is well known that the game ends with probability 1, and that the mean time for the game to end is infinite.

I am interested in the following question: What is the (asymptotic) probability that the game is not yet over after $$n$$ flips?

From a heuristic argument, I'm fairly certain that the answer is $$\theta(1/\sqrt{n})$$. From simulation, it appears that the answer is about $$0.8/\sqrt{n}$$.

I'd like to know the exact answer, and I'd like to know how to derive it analytically. At least, I'd like to know how to prove that the probability is $$\theta(1/\sqrt{n})$$. I'm guessing the proof involves a martingale, but I can't find it myself.

The exact probability that the game has not ended after the $$\ n^\text{th}\$$ toss is $$\frac{\pmatrix{n\\\left\lfloor\frac{n}{2}\right\rfloor}}{2^n}\sim\sqrt{\frac{2}{\pi n}}\ .$$ The proof of the first expression turns out to be more straightforward than I had initially expected. The asymptotic approximation follows from the well known asymptotic expressions for the central binomial coefficients: \begin{align} {2n\choose n}&\sim\frac{4^n}{\sqrt{\pi n}}=2^{2n}\sqrt{\frac{2}{2n\pi}}\\ {2n+1\choose n}&={2n+1\choose n+1}\sim\frac{2^{2n+1}}{\sqrt{\pi(n+1)}}\\ &=2^{2n+1}\sqrt{\frac{2}{\pi(2n+1)}}\sqrt{1-\frac{1}{2n+2}}\ . \end{align} For $$\ i\ge1\$$ let $$\ p_{in}\$$ be the probability that the player has $$\ i\$$ dollars after the $$\ n^\text{th}\$$ toss, and let $$\ p_{0n}\$$ be the probability the game ends on or before the $$\ n^\text{th}\$$ toss. Then \begin{align} p_{n+1\,n}&=\frac{1}{2^n}\ ,\\ p_{n\,n}&=0\ ,\\ p_{0\,n}&= p_{0\,n-1}+\frac{p_{1\,n-1}}{2}\ ,\\ p_{1\,n}&= \frac{p_{2\,n-1}}{2}\ , \text{ and}\\ p_{i\,n}&= \frac{p_{i+1\,n-1}+p_{i-1\,n-1}}{2}\ \text{ for }\ i\ge2\ . \end{align} To simplify the calculation, let $$\ T_{nj}=2^{n+j}p_{n+1-j\,n+j}\$$ for $$\ 0\le j. Then \begin{align} T_{n0}&=1\ ,\\ T_{11}&=1\ ,\text{ and}\\ T_{nj}&=T_{n\,j-1}+T_{n-1\,j}\ \text{ for }\ 1\le j\le n\ . \end{align} It follows from the last of these identities that $$T_{nk}=\sum_{j=0}^kT_{n-1\,j}\ .$$ The numbers $$\ T_{nj}\$$ are the entries in Catalan's triangle. The numbers $$\ T_{nn}\$$ along the diagonal are the Catalan numbers, $$T_{nn}=\frac{2n\choose n}{n+1}\ ,$$ from which we obtain \begin{align} p_{1\,2n}&=\frac{T_{nn}}{2^{2n}}\\ &= \frac{2n\choose n}{(n+1)2^{2n}}\ . \end{align} From the recurrence for $$\ p_{in}\$$ we also get $$\ p_{1\,2n+1}=p_{2\,2n}=0\$$ and \begin{align} p_{0\,2n}&=p_{0\,2n-1}\\ &=p_{0\,2n-2}+\frac{p_{1\,2n-2}}{2}\\ &= p_{0\,2n-2}+\frac{2n-2\choose n-1}{n2^{2n-1}} \end{align} It can be verified by induction that the solution of this recurrence is \begin{align} p_{0\,2n}&=1-\frac{2n\choose n}{2^{2n}}\\ &=1-\frac{2n-1\choose n-1}{2^{2n-1}}\\ &=p_{0\,2n-1}\ . \end{align} Now $$\ p_{0n}\$$ is the probability that after the $$\ n^\text{th}\$$ toss the game has ended, so the probability that the game has not ended after the $$\ n^\text{th}\$$ toss is $$1-p_{0\,n}= \frac{\pmatrix{n\\\left\lfloor\frac{n}{2}\right\rfloor}}{2^n}\ ,$$ as stated above.