Success probability that makes eight of twelve most likely

A random variable $Y$ is binomial with $12$ trials and success probability $p\in(0,1)$. For which $p$ is $\mathbb{P}(Y=8)$ as large as possible?

$p=\tfrac23$. Maximise $\log\big[p^8(1-p)^4\big]=8\log p+4\log(1-p)$: the derivative $\tfrac8p-\tfrac4{1-p}$ vanishes at $p=\tfrac{8}{12}$. In general $\mathbb{P}(Y=k)$ is maximised at the sample proportion $p=k/n$ (the maximum-likelihood estimate).