# Rough bound for sum $\binom{3n}{0}+\binom{3n}{1}+\cdots+\binom{3n}{n-1}$

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.

Here's an outline of a proof. We have that $\binom{3n}{0}+\binom{3n}{1}+\cdots+\binom{3n}{3n} = 2^{3n}$. Split the sum into thirds. The middle third is clearly larger than the other two, which, being equal by symmetry, must therefore each account for less than a third of the total.
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 $n=2$.