Chernoff bounds and Monte Carlo algorithms
Let's suppose that we have a polynomial time randomized algorithm $A$ for a decision problem $\Pi$ with the property that
$$
\mathrm{Pr}[A \text{ is right}] &gt; 1/2
$$
for all inputs.  (In particular, independent of the input length $n$.)

We can use Chernoff bounds to show something stronger, namely that for any fixed $c$, there is a polynomial time, randomized algorithm for $\Pi$ that is correct with probability at least $1-n^{-c}$.

The algorithm is very simple: On an input of length $n$, we run $A$ a polynomial number of times $N$ (to be determined below) and then report the majority answer.  Call this algorithm $A'$.

What we need is a bound on 
$$
\mathrm{Pr}[A' \text{ is wrong}]
$$
in terms of the number of repetitions $N$.  The hypothesis about $A$ implies that there is an $\epsilon &gt; 0$ such that 
$$
\mathrm{Pr}[A \text{ is right}] \ge 1/2 + \epsilon
$$
Since the $N$ runs of $A$ are independent, for any $k\in [N]$ we have
$$
\mathrm{Pr}[k \text{ of } N \text{ runs of } A \text{ are right}]\ge Pr[X = k]
$$
where $X$ is a binomial random variable with parameters $N$ and $1/2+\epsilon$.  Since the algorithm $A'$ is wrong exactly when at most $N/2$ of the runs of $A$ are right, we obtain an upper bound on the failure probability by a "tail estimate" bounding 
$$
\mathrm{Pr}[X \le N/2]
$$
This is a prototypical time to use a Chernoff bound, since a binomial r.v. is the sum of $N$ independent random variables.  One instance of the Chernoff bound says that if $Y$ is the sum of $N$ independent r.v.'s with support in $[0,1]$ and $t &gt; 0$, then 
$$
\mathrm{Pr}[Y &lt; E[Y] - t] \le e^{-2t^2/N}
$$
Thus, since $E[X] = N/2 + N\epsilon$, we take $t = N\epsilon$ to obtain 
$$
\mathrm{Pr}[A' \text{ is wrong}]\le e^{-2\epsilon^2 N}
$$
So we can take $N = (1/2)\epsilon^{-2}c\log n$, and we get the claimed error probability for $A'$.