Is $\binom{52}{n}\cdot\binom{52-n}{n}$ maximised by $n=\frac{52}{3}$? If so why?

Let $a_n$ be your product of binomial coefficients. It is clear that $a_n$ starts out small.

We calculate $\dfrac{a_{n+1}}{a_n}$. Looks bad at first, but there is a tremendous amount of cancellation, and after a while we find that $$\frac{a_{n+1}}{a_n}=\frac{(52-2n)(52-2n-1)}{(n+1)^2}.$$

We can now use a little algebra to find out where the function $$f(t)=\frac{(52-2t)(52-2t-1)}{(t+1)^2}$$ is greater than $1$, and where it is less than $1$. This turns quickly into a problem about a quadratic. We get equality at roughly $16.8$. So for $n\le 16$, $a_{n+1}$ is "better" than $a_n$. The maximum value of $a_n$ is therefore at $n=17$.

Details: We have $$a_n=\frac{52!}{n!(52-n)!}\frac{(52-n)!}{(n!)(52-2n)!}=\frac{52!}{n!n! (52-2n)!}.$$ Mechanically, it follows that $$a_{n+1}= \frac{52!}{(n+1)!(n+1)!(50-2n)!}.$$ Divide the second by the first. We get $$\frac{a_{n+1}}{a_n}=\frac{n!n!(52-2n)!}{(n+1)!(n+1)!(50-2n)!}.$$ Now $(n+1)!=n!(n+1)$ and $(52-2n)!=(52-2n)(52-2n-1)(50-2n)!$, and we have our simplified ration.

Let us define $\binom{n}{k}$ for non-integer $n$ and $k$ by:

$$ \binom{n}{k} = \frac{\Gamma(n+1)}{\Gamma(n-k+1)\Gamma(k+1)} $$

where $\Gamma(n) = \int_0^\infty x^{n-1} e^{-x} dx$ is the analytical continuation of the factorial (with $\Gamma(n+1) = n!$ for integer $n$). We must do this because the factorial function is not actually defined on the integers. One way to define it is the Gamma function, which matches up with the factorial for integer arguments. Now that we have the Gamma function, our problem reduces to calclus.

Then we have that:

$$ \binom{52}{n} \binom{52-n}{n} $$ $$ \frac{\Gamma(53)}{\Gamma(53-n)\Gamma(n+1)} \cdot \frac{\Gamma(53-n)}{\Gamma(53-2n)\Gamma(n+1)}$$ $$ \frac{\Gamma(53)^2}{\Gamma(53-2n)\Gamma(n+1)^2}$$

The numerator is a large number ($52!^2$). So let us worry about the denominator instead. The larger the denominator is, the smaller the fraction will be. To maximize the fraction, we should minimize the denominator. So:

$$ 0 = \frac{d}{dn} \Gamma(53-2n)\Gamma(n+1)^2 $$ $$ 0 = -2\Gamma^\prime(53-2n)\Gamma(n+1)^2 + 2\Gamma^\prime(n+1)\Gamma(n+1)\Gamma(53-2n) $$ $$ \Gamma^\prime(53-2n)\Gamma(n+1) = \Gamma^\prime(n+1)\Gamma(53-2n) $$

Which looks extremely ugly. One obvious way they could be equal is if $n+1 = 53-2n$, because then, by simple substitution, the two sides are equal. Solving for $n$, we get $n = \frac{52}{3}$.