Bernoulli theorem and frequency interpretation of Bernoulli trials We have a Bernoulli theorem/Law of Large numbers and Chebyshev's inequality almost saying the same thing but I'll state the theorem in relation to bernoulli trials
Let probablility of an event in a single Bernoulli trial be $p$. Then for $n$ such trials, we have
\begin{equation}
P\left\{\left|\frac{k}{n}-p\right|\gt\epsilon\right\}\lt \frac{pq}{n\epsilon^2}
\end{equation} for given $\epsilon \gt 0$ and $p+q=1$, where $k$ is the no. of successful trials. 
On the other hand, we have $P\{k=np\}=0$ as $n\rightarrow\infty$. (Take the case when $np$ is an integer)
Now in the former, one can tighten the bound for arbitrarily small $\epsilon$ by taking $n$ very large, in which case $k$ will depend on $n$. On the other hand, the latter is also very logical since hitting a particular number in a sea of numbers should be close to improbable.
The general explanation to this conundrum is given as that the bound is an interval. In that case as the bound $\epsilon$ gets tighter the neigbouring values of $k$ in the vicinity of $k=np$ will rise to satisfy the frequency interpretation that $k\sim np$ for large $n$. This is the basis of DeMoivre-Laplace approximation as well. 
But I have a very uneasy feeling about it. I just want a illustration with an explanation that this is true.
 A: Using $\{X_i\}$ as the iid Bernoulli random variables, we can write $K=\sum_{i=1}^n X_i$ and indeed we have for every $\epsilon>0$ and every positive integer $n$: 
$$ P\left[\left|\frac{1}{n}\sum_{i=1}^n X_i - p\right| \geq \epsilon\right] \leq \frac{p(1-p)}{n\epsilon^2} \quad (\mbox{Chebyshev})$$
As BGM comments, the event $\{\left|\frac{1}{n}\sum_{i=1}^n X_i - p\right| \geq \epsilon\}$ says that $\sum_{i=1}^n X_i$ lies in an interval of size $2n\epsilon$.  Thus,
$$ P\left[-n\epsilon \leq \sum_{i=1}^n X_i \leq n\epsilon\right] \leq \frac{p(1-p)}{n\epsilon^2}  $$
and for any $\epsilon>0$, this interval size $2n\epsilon$ grows infinitely large with increasing $n$. 
Now if we want to shrink $\epsilon$ by a factor of 10 while maintaining the same right-hand-side in the Chebyshev bound, we must increase $n$ by a factor of 100.  This increases the interval of size $2n\epsilon$ by a factor of 10. 

To plug specific numbers, suppose $p=1/2$ and $\epsilon=1/1000$.  We have: 
$$ P\left[\left|\frac{1}{n}\sum_{i=1}^n X_i - p\right| \geq \epsilon\right] \leq \frac{10^6}{4n} $$
In order for the right-hand-side to be less than or equal to $1/1000$, we need $10^6/(4n) \leq 1/1000$, which means $n\geq 25 \times 10^7$ and the interval of size $2n\epsilon$ contains about $250000$ integers!  This includes the integer $np = \lceil n/2\rceil$, but also many many more integers to choose from.  With so many choices, it is easy to see that the probability of getting $K=\lceil n/2 \rceil$ is very small (without contradicting the Chebyshev inequality which says $K$ is one of those 250000 integers in the size-$2n\epsilon$ interval with high probability). 
