If random variables are drawn from an identical distribution, why doesn't this guarantee they are independent? Having read a little about exchangeability, I went back to thinking about the iid condition required for the central limit theorem. It struck me that if two random variables are drawn from an identical distribution, the occurrence of one event does not make it more or less probable for the occurrence of another event. I have no doubts that I am wrong and there is a reason why both independence and identical distribution are required. I just don't know why!
 A: 
If random variables are drawn from an identical distribution, why doesn't this guarantee that they are independent?

Since you don't specify how the random variables are drawn, the question has no meaning.  It is the manner of drawing that is important.  Consider a neoclassical example of an urn with one ball marked $0$ and one ball marked $1$.  The random variable $X$ is the number on the ball drawn from the urn, and is a Bernoulli random variable with parameter $p = P\{X = 1\} = \frac{1}{2}$.  Now let $X_1$ denote the number on the first ball drawn and $X_2$ the number on the second ball drawn.


*

*Case I: drawing with replacement  There are 4 equally likely outcomes of the experiment and they can be written as $00, 01, 10, 11$.
$X_1$ and $X_2$ are both 
Bernoulli$\left(\frac{1}{2}\right)$ random variables and are independent.  

*Case II: drawing without replacement Now there are only two 
equally likely outcomes $01, 10$ but $X_1$ and $X_2$ clearly are
Bernoulli$\left(\frac{1}{2}\right)$ random variables just as before, 
and just as clearly are not independent.


Thus, just getting random variables with identical distribution does
not by any means guarantee that they are independent.
