Identifying binomial distribution for finding variance 
If a variable $x$ takes values $0,1,2,....n$ with frequencies equal to the binomial coefficients $\binom {12}0,\binom {12}1,\binom {12}2,.....\binom {12}{12}$, then variance of distribution is

\begin{array}{|c|c|c|c|}
\hline
x& 0 & 1 & 2&......&12 \\ \hline
f &\binom {12}0 &\binom {12}1 &\binom {12}2&......&\binom {12}{12}\\ \hline
\end{array}
It is solved in my reference as
$$
\sigma^2=n/4=12/4=3
$$
as if it is a binomial distribution with $p=q=1/2$.
I understand it must be a binomial distribution but where do we have the clue that the probability of success in each trial is $1/2$ ?
My thinking says irrespective of the probability of success the frequency of each case is the above binomial coefficients. So where am I thinking wrong about it ?
 A: This is not a 'binomial distribution' as stated. You are given an ordinary frequency distribution and you are asked to calculate its variance, i.e. variance of the dataset $\{x_0,x_1,\ldots,x_{12}\}$ where $x_i=i$ is the variable of interest with corresponding frequency $f_i=\binom{12}{i}$, $i=0,1,\ldots,12$. So you can just calculate variance like you do for a grouped data. 
But you can think of a certain probability if you consider the relative frequency $$f(x)=\frac{f_x}{\sum_{x=0}^{12}f_x}=\frac1{2^{12}}\binom{12}{x}\,.$$ 
This as you know is the probability mass function of a binomial distribution with parameter $\left(12,\frac12\right)$ evaluated at $x$ when $x\in\{0,1,\ldots,12\}$. It will be immediate that the required variance is just the variance of this binomial distribution once you write down the expression of variance for grouped data.
A: Think about coin flips. If a biased coin has probability $p$ of landing heads, and $X$ is a random variable (note that $X$ is binomially distributed by design) equal to the number of heads in $12$ flips, then the probability distribution of $X$ is 
$$P(X = k) = \binom{12}{k}p^k(1-p)^{12-k}$$
For each $k$, this is proportional to $\binom{12}{k}$ iff $p=\frac{1}{2}$. 
