I am not a mathematician, so I don't quite understand how comes that a variance of some discrete probability distribution could exceed the variance of the discrete uniform distribution. I thought that variance of any distribution will be capped by the variance of the flat distribution.
My case:
I have the following probability distribution for the discrete random variable on the [0,4] interval:
0.66$\hspace{0.5cm}$0.16$\hspace{0.5cm}$0.01$\hspace{0.5cm}$0.01$\hspace{0.5cm}$0.16
I compute the variance of this distribution using:
$$
\text{Var}(X) = \sum_{i=1}^n p_i\,\cdot\,(x_i - \mu)^2
$$
and I get 2.1275, while the variance for the discrete uniform distribution, according to this formula
$$
\text{Var}(X) = \frac{n^2 - 1}{12}, \quad \text{where }n=b-a+1
$$
should be 2.0 on the same interval, i.e., for $b-a = 4$.
This bugs me a lot. Please tell me what I am missing. Thank you for your time.