How to maximize $\left({a+b \choose a} 2^{-a-b}\right)$? How can you maximize $\left({a+b \choose a} 2^{-a-b}\right)$ assuming, $a,b \geq 0$ and $0< (a+b) \leq n$, where all the variables are non-negative integers?  Is the maximum when $a=b=n/2$, assuming $n$ is even? 
 A: The expression can be viewed as the probaility of having exactly $a$ heads in $a+b$ fair coin tosses, so intuitively, this should be maximized if $a=b$. Moreover, the more coin tosses we make, the less likely it will become to obtain exactly (rounded) half of them heads.
Formally, with $f(a,b)={a+b\choose a}2^{-a-b}$ we have
$$ \frac{f(a+1,b-1)}{f(a,b)}=\frac{a+b\choose a+1}{a+b\choose a}=\frac{a!b!}{(a+1)!(b-1)!}=\frac{b}{a+1}$$
so that the maximum is certainly not assumed at a pair $(a,b)$ with $b>a+1$. By symmetry, i.e. $f(a,b)=f(b,a)$, we can also exclude $a>b+1$. Thus we only have to maximize among $f(a,a)$ with $0<a\le\frac n2$ and $f(a,a+1)$ with $0\le a\le\frac{n-1}2$.
From the additive recrusion for binmoials,
$$ \begin{align}{2a+2\choose a+1}&={2a+1\choose a}+{2a+1\choose a+1}\\&={2a\choose a-1}+2{2a\choose a}+{2a\choose a+1}\\&=\left(\frac{a-1}{a}+2+ \frac{a-1}{a}\right){2a\choose a}\\&<4{2a\choose a}\end{align}$$
so that we see that the maximuum among all $f(a,a)$ is $f(1,1)=\frac12$.
From ${2a+2\choose a+1}=2{2a+1\choose a}$ we see that $f(a,a+1)=f(a+1,a+1)$, so the maximum of these values is $f(0,1)=f(1,1)=\frac12$.
To summarize: We have $f(0,1)=f(1,0)=f(1,1)=\frac12$ and $f(a,b)<\frac 12$ for all other choices of $a,b$.
