Binomial distribution as a function of $n$ and $k$

Question

As we know, the probability that we get exactly $k$ successes after $n$ experiments is: $\mathbb{P}(S_n = k) = C_n^k\cdot p^k(1-p)^{n-k}$, where $p$ is the probability to get a success in $i^{th}$ experiment.
Let us look at $\mathbb{P}(S_n = k)$ as a function of two variables; $n$ and $k$. Then:
1. If we fix $k$, for what values of $n$, the probability is maximal?
2. If we fix $n$, for what values of $k$, the probability is maximal?
Logically, the answer to the second one must be $\left\lceil\frac{n}{2}\right\rceil$ if $p>1/2$ and $\left\lfloor\frac{n}{2}\right\rfloor$ if $p<1/2$. I can't come up with a proof. Obviously differentiation is not an option. I tried using Stirling's formula or taking $\ln$ of both sides, but still couldn't get anything.

Answer 1

Your answer to the second part is wrong, it’s either the floor or ceiling of $pn$. The proof of this is to look at successive ratios between terms. Notice that it increases up to $pn$, then decreases afterwards, and hence the maximum must be around there.

On the other hand, if you fix $k$ and want to maximise it for $n$, the method is exactly the same.