If I throw a fair dice $12$ times, the expected number of $6$ is $2$ i.e $6$ is expected to happen $2$ times when the dice is thrown $12$ times. But the probability of getting $6$ exactly $2$ times is ${12}\choose{2}$$(1/6)^{2} (5/6)^{10}$ which is less than $1$.
I am a first reader of probability. I am quite curious to know why this is happening?
you are expecting $6$ appearing two times . On the other hand you are telling that the possibility of appearing 6 , two times is very low. How these two fact go simultaneously? I can not understand this.
I feel that the probability of getting six, two times should have been close to $1$. Can anyone please explain me where I am going wrong?
Maybe it helps you understand what is happening, if you consider an extreme case of the same phenomenon: the probability of a random variable being equal to its expected value can even be zero. Just think of
$$ P(X=1)=P(X=-1)=0.5,$$
then
$$ E(X)=-1\cdot P(X=-1)+1\cdot P(X=1)=-0.5+0.5=0,$$
and hence
$$ P(X=E(X))=P(X=0)=0.$$
Now, let's take a look at your particular example of a binomial distribution, that is you have $n\in\Bbb N$ and $p\in(0,1)$ such that
$$P(X_n=k)={}^nC_k p^k(1-p)^{n-k}, \quad k\in\{0,\ldots,n\}.$$
You can actually show that for all $k$, this expression becomes arbitrarily small if you increase $n$.
And finally, a very pragmatic suggestion: If your problem is mostly about intuition, just go ahead and do the experiment yourself. Take a coin, toss it ten times, see how many times you got head, write it down, repeat. You'll see that exactly five heads will be rather rare.