Sum of variances of multinomial distribution. I've k fair coins, and I would like to know the number of heads obtained in $n$ trials. But that is simple binomial distribution. But if I want to find out how much it varies from binomial expectation, them it becomes a multinomial distribution for the number of heads from 0 to k.
 $$ \text{pdf} = \frac{n!}{x_0!\cdots x_k!}p_0^{x_0}\cdots p_k^{x_k} $$
Where:
$$ p_i =  {k \choose i}\frac{1}{2^k}$$
I am interested in:
$$ \text{Var}(\vec{X}) = \sum_{i=0}^k E(X_i - E[\vec{X}]_i)^2$$
Since we have:
$$ \text{Var}(X_i) = E(X_i - E[X_i])^2 = np_i(1-p_i)$$
The wrong derivation is as follows, how do I correct this. This I think is wrong due to dependency or number of degrees problem. 
$$ \text{Var}(\vec{X}) = \sum_{i=0}^k\text{Var}(X_i) = \sum_{i=0}^k np_i(1-p_i)$$
on substitution we get:
$$ \text{Var}(\vec{X}) = n(1-{2k \choose k}) \approx n(1-\frac{1}{\sqrt{\pi~k}})$$
How do I correct this? Experimental data (for k=15) seem to differ a bit. In the following graph RMS is defined as:
$$ \text{RMS} = \frac{1}{n}\sqrt{\text{Var}(\vec{X})}$$

The IJulia code used to generate this is: here

There is nothing wrong with the derivation, only problem with experimental setup. The experiment is calculating expected value of standard deviation and not of variance. 
Change
RMSExptAvg(S, n) = mean([RMSExpt(S, n) for i in 1:10000]))

to this:
RMSExptAvg(S, n) = sqrt(mean([RMSExpt(S, n)^2 for i in 1:10000]))

Then equations match up exactly.
Getting closed form equation for expectation of standard deviation is pretty difficult.
