TITLE: Rough bound for sum $\binom{3n}{0}+\binom{3n}{1}+\cdots+\binom{3n}{n-1}$
QUESTION [8 upvotes]: Is it true that
$$\frac{\dbinom{3n}{0}+\dbinom{3n}{1}+\cdots+\dbinom{3n}{n-1}}{2^{3n}}<\frac13$$ for all positive integers $n$?
I've plotted the first few values of $n$ and noticed that the left-hand side decreases rapidly, so I'm quite sure that it's true. But how to prove it? Induction doesn't looks like helping much here.

REPLY [7 votes]: The Binomial Distribution $2^{-3n}\binom{3n}{k}$ has mean $\frac32n$ and variance $\frac34n$. Thus, by Chebyshev's Inequality, we have
$$
P\left[|x-\tfrac32n|\ge a\right]\le\frac{\frac34n}{a^2}\tag{1}
$$
Letting $a=\frac12n+1$ yields
$$
\begin{align}
P\left[|x-\tfrac32n|\ge\tfrac12n+1\right]
&\le\frac{\frac34n}{(\frac12n+1)^2}\\
&=\frac{3n}{(n+2)^2}\tag{2}
\end{align}
$$
By the symmetry of the binomial distribution, we get
$$
\begin{align}
P\left[x\le n-1\right]
&=P\left[x-\tfrac32n\le-\tfrac12n-1\right]\\
&\le\bbox[5px,border:2px solid #C0A000]{\frac{3n}{2(n+2)^2}}\tag{3}
\end{align}
$$
which is less than $\frac13$ for all $n$. In fact, it reaches a maximum of $\frac3{16}$ at $n=2$.