Here's a question I'm struggling with:
A man is standing on $0 \in \mathbb{R}$. He tosses $n$ coins. For H he takes a right step and for T he takes a left step. What is the probability that at the end of the process he's standing on a non-negative integer?
Here's my attempt.
Let $P(n\text{ is odd}) = p$ and $P(n\text{ is even}) = 1-p$
If $n$ is odd we know that he can't go back to zero. The number of paths that ends at $\mathbb{R}^+$ is equal to those that ends at $\mathbb{R}^-$ so the probability in this case is $\dfrac{2^n}{2}$.
If $n$ is even we need to cound the number of string where $H \ge T$ and so the probability is $\frac{\sum_{k=n/2}^{n}\binom{n}{k}}{2^n}$
and so in general we get $(1-p)\cdot \frac{2^n}{2} + p\cdot \frac{\sum_{k=n/2}^{n}\binom{n}{k}}{2^n}$
Am I right? Thanks!