I'll interpret "most likely" as Maximum Likelihood Estimation, that is, given some observations, what is the value of $p$ that makes the probability of those observations the largest. For example, if we observe something occurring $0$ out of $n$ times, the maximum likelihood estimate for $p$ is $0$, because that gives our observations a probability of $1$. This is only unreasonable if you have some prior information about $p$ (such as "it's probably around $0.5$, but maybe a little higher or lower" or "it's very close to either $1$ or $0$, but I don't know which").

Suppose we run $n$ trials, and find that the event occurred $n$ times, and we assume that each trial is independent with the same probability $p$ of success. Then the probability of our observation is
$$P(p) = {n \choose m} p^m (1-p)^{n-m},$$
since the probability of each of the $m$ successes is $p$, the probability of the $n-m$ failures is $1-p$, and there are ${n\choose m}$ ways for $m$ of the $n$ trials to be the successful trials. Assuming $m$ and $n-m$ are both nonzero, this probability vanishes when $p=0$ or $1$, so the maximizing value of $p$ will be somewhere in between. Then we can find this maximizing value of $p$ by taking the derivative of $P$ and setting the result equal to $0$. We can take the derivative:
$$P'(p) = {n\choose m} \left[mp^{m-1}(1-p)^{n-m} - (n-m)p^m(1-p)^{n-m-1}\right]$$
$$ = {n\choose m} \left[m(1-p) - (n-m)p\right]p^{m-1}(1-p)^{n-m-1}$$
This vanishes when $m(1-p)-(n-m)p=0$, i.e. when $m = np$ or $p = m/n$.

So if you have no prior information about what $p$ should be, but you observe $m$ successes in $n$ independent trials, the value of $p$ that best matches your observation is $p=m/n$.