# Gambler with infinite bankroll reaching his target

Suppose a gambler with infinite bankroll has a target of winning 10 dollars. He wins/loses $\$1$with probabilities$0.48=p$and$0.52=q$respectively. What is the probability that he meets the target? The answer using the usual methods is$(p/q)^n = (12/13)^{10}.$By a rather devious process, I have arrived at a combinatorial formula, $$\sum_{k=n}^{\infty}\frac{n}{k}{2k-n-1\choose k-n}p^kq^{(k-n)},$$ I get the same results, but can it be proved that the results will be identical? - ## 1 Answer Here's a probabilistic argument. Let$T$be the hitting time of the state$n$. Then a simple application of the hitting time theorem shows that, for$k\geq n$, $$\frac{n}{k}{2k-n-1\choose k-n}p^kq^{(k-n)}=\mathbb{P}(T=2k-n).\tag1$$ Therefore $$\sum_{k=n}^{\infty}\frac{n}{k}{2k-n-1\choose k-n}p^kq^{(k-n)}=\mathbb{P}(T<\infty).\tag2$$ As for an analytic argument, use a change of variables and simple properties of binomial coefficients to rewrite the sum on the left hand side of (2) as $$\sum_{x=0}^{\infty}\frac{n}{2x+n}{2x+n\choose x}p^{x+n}q^x.\tag3$$ Using formula (5.70) on page 203 of Concrete Mathematics by Graham, Knuth, and Patashnik, the sum in (3) can be written as $$(p\,{\cal B}_2(pq))^n=\left(1-\sqrt{1-4pq}\over 2q\right)^n.$$ For$0\leq p\leq 1/2$and$q=1-p$, this reduces to$(p/q)^n\$.

-
Thanks a bunch ! – true blue anil Sep 27 '12 at 4:02