Is covariance of two random variables from the same distribution the same as the variance of this distribution? Sorry that this might be a very simple question, but I got confused: say we have a Binomial distribution $Bin(n, p)$, and two random variables, $X$ and $Y$, drawn from it. 
Is the covariance between $X$ and $Y$ as simple as $np(1-p)$ (i.e. the variance of this Binomial distribution)?
 A: The covariance of two random variables with the identical variance $\sigma^2$ (note, no requirement that the distributions be identical or that they be binomial, etc) always has value in the interval $[-\sigma^2, \sigma^2]$. So, Yes, the covariance can have value $\sigma^2$ but this happens only when the random variables are equal, that is, $X$ always has the same value as $Y$ for every possible outcome of the experiment.
A: Consider the simple experiment of rolling two fair six-sided dice. The outcomes on the two dice may reasonably be taken to be independent.
Let's call it a success if a die shows 5 or 6 and a failure otherwise. That is, let X and Y be 1 if the first and second die (respectively) show a 5 or 6 and 0 otherwise, so that they form an independent, identically distributed pair of Bernoulli random variables (i.e. just as in your question, but for the special case n=1)
While their distribution is the same, $\text{cov}(X,Y) = 0$ (indeed, clearly they're independent). 
[You can easily write their joint distribution and explicitly compute their covariance if you wish.]
