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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?

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up vote 1 down vote accepted

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 $(p/q)^n$.

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