Let $\operatorname{Binom}(n, p)$ be $n$ trials with probability of success $p$. I want to find
$$\lim\limits_{n \to \infty} P(\operatorname{Binom}(n, p) \geq n/2)$$
I didn't know how to do this with the binomial distribution itself, so I tried transforming it into a random walk. That is, if we have $A_i = 1$ with probability $p$ and $A_i = -1$ with probability $1-p$, this is equivalent to
$$\lim\limits_{n \to \infty} P\left(\sum\limits_{i = 1}^n A_i \geq 0\right)$$
but I wasn't sure where to go from here either.
Best Answer
Note that $\operatorname{Var}(\operatorname{Binom}(n,p))=np(1-p)$, and so you can use Chebyshev's inequality to show that $$P\left(\left|\frac{\operatorname{Binom}(n,p)}{n}-p\right|>\epsilon\right)\to 0$$ for any $\epsilon>0$.
This immediately gives the answer $0$ for any $p<1/2$ and $1$ for any $p>1/2$. For $p=1/2$ you also need to check that $P(\operatorname{Binom}(n,1/2)=n/2)\to 0$; one way to do this is by Stirling's approximation.