Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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 $(p/q)^n$.

share|cite|improve this answer
Thanks a bunch ! – true blue anil Sep 27 '12 at 4:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.