Result of Bernoulli trials being twice the expectancy? Given a probability $0 < p < 0.5$ for success per trial with $n$ Bernoulli trials, what are the odds for having succeeded in at least $2np$ experiments?
 A: The large deviations principle alluded to in the comments is as follows. Consider the number $S_n=X_1+\cdots+X_n$ of successes during the $n$ first trials. Then $(X_n)_{n\geqslant1}$ is i.i.d. with Bernoulli distribution with parameter $p$, in particular, for every $t$, $\mathbb E(\mathrm e^{tX_k})=p\mathrm e^t+1-p$. 
Let $x\gt p$ and $A_n(x)=[S_n\geqslant nx]$. Then, almost surely, $\mathrm e^{tnx}\mathbf 1_{A_n(x)}\leqslant\mathrm e^{tS_n}$ for every $t\geqslant0$, hence, integrating both sides of this inequality, one sees that
$$
\mathbb P(A_n(x))\leqslant\mathrm e^{-tnx}\mathbb E(\mathrm e^{tS_n})=\mathrm e^{-tnx}\mathbb E(\mathrm e^{tX_1})^n=\mathrm e^{-n\Lambda(t,x)},$$
with
$$
\Lambda(t,x)=tx-\log\mathbb E(\mathrm e^{tX_1})=tx-\log(p\mathrm e^t+1-p).
$$
The most interesting upper bound this technique can yield is the one such that $\Lambda(t,x)$ is maximal, that is, when $t=t(x)$ with $\mathrm e^{t(x)}=(1-p)x/(p(1-x))$. When $p\lt x\lt1$, $t(x)$ is finite and positive and
$\Lambda(t(x),x)=\Lambda(x)$ with
$$
\Lambda(x)=x\log\left(\frac{x}p\right)+(1-x)\log\left(\frac{1-x}{1-p}\right).
$$
It happens that, for each $p\lt x\lt1$, $\Lambda(x)$ indicates the exact order of exponential convergence of $\mathbb P(A_n(x))$ when $n\to\infty$, that is,
$$
\lim\limits_{n\to\infty}\frac1n\log\mathbb P(S_n\geqslant nx)=-\Lambda(x).
$$
In your case, $0\lt p\lt\frac12$ and $x=2p$ hence, for every $n$,
$$
\mathbb P(S_n\geqslant 2np)\leqslant\exp\left(-n\left(2p\log2+(1-2p)\log\left(\frac{1-2p}{1-p}\right)\right)\right),
$$
and, when $n\to\infty$,
$$
\mathbb P(S_n\geqslant 2np)=\exp\left(-n\left(2p\log2+(1-2p)\log\left(\frac{1-2p}{1-p}\right)\right)+o(n)\right).
$$
The argument briefly sketched above is due to Harald Cramér and can be adapted to incredibly more general situations.
A: You have a variable following the binomial distribution $B(n, p)$, with $E[X] = np$ and $Var[X] = np(1-p)$. Generally speaking, the probability that $X \geq 2np$ is small (actually very small).
By Markov's inequality, let $t = 2np$,
$$P[X\geq t] \leq \frac{E[X]}{t} = \frac{np}{2np} = 0.5.$$
But this bound is not good.
By Chebyshev's inequality, let $t = np$,
$$ P[|X-E[X]|\geq t] \leq \frac{Var[X]}{t^2} = \frac{np(1-p)}{(np)^2} = \frac{1-p}{np}.$$
By Hoeffding's inequality, let $\epsilon = p$,
$$P[X \geq (p+\epsilon)n] \leq \exp(-2\epsilon^2n) = \exp(-2p^2 n).$$
Hoeffding's inequality gives the best bound: the probability goes down exponentially as the number of trials grows. I believe the Chernoff bound gives similar result.
http://en.wikipedia.org/wiki/Concentration_inequality
http://en.wikipedia.org/wiki/Hoeffding%27s_inequality
A: Appendix A of the text Error-Correcting Codes by W. W. Peterson and E. J. Weldon, Jr.
(2nd edition, MIT Press, 1972) credits the following to notes transcribed from the
lectures of Claude Shannon at a MIT seminar.
If $\lambda > p$, then
$$\binom{n}{\lambda n}p^{\lambda n}(1-p)^{(1-\lambda)n}
< \sum_{i=\lambda n}^n \binom{n}{i}p^i(1-p)^{n-i} 
< \frac{\lambda(1-p)}{\lambda - p}\binom{n}{\lambda n}p^{\lambda n}(1-p)^{(1-\lambda) n},$$
and
$$\sum_{i=\lambda n}^n \binom{n}{i}p^i(1-p)^{n-i} 
\leq \left(\frac{p}{\lambda}\right)^{\lambda n}\left(\frac{1-p}{1-\lambda}\right)^{(1-\lambda)n}$$
Here, $\lambda = 2p > p$, and so we have
$$\begin{align*}
\sum_{i=2pn}^n \binom{n}{i}p^i(1-p)^{n-i} 
&\leq 2^{-2np}\left(\frac{1-p}{1-2p}\right)^{(1-2p)n}\\
&= \exp\left(-n\left(2p\ln 2+(1-2p)\ln\left(\frac{1-2p}{1-p}\right)\right)\right)
\end{align*}$$
which is essentially the result already given by @did 
(except without the $o(n)$ term since the above is an upper bound).
The proof of these inequalities uses the upper bound
$$\binom{n}{\lambda n} < \frac{1}{\sqrt{2\pi n \lambda(1-\lambda)}}
\lambda^{-\lambda n}(1-\lambda)^{-(1-\lambda)n}$$
which is itself proved using Stirling's approximation for $n!$
