Standard deviation of a baised $d$-sided coin I know that that standard deviation of a noisy bit (a biased coin with probability distribution $\{ p, 1-p \}$ ) is given by 
$$ \sigma = \sqrt{p(1-p)} $$
What is then a measure of the standard deviation for the d-dimensional variant? e.g. for $d=3$ we have state space $$\{p_1, p_2, 1-p_1-p_2\}$$
Many thanks!
 A: Answer in development
Let a $d$-sided coin have sides labeled $X\in\{1,2,3,\ldots,d\}$, and let the state space be
$$\{p_1,p_2,p_3,\ldots,p_d\}$$
this is, $p_i$ is the probability of throwing the side labeled $i$. Note that $$\sum_{i=1}^dp_i=1$$
Then, the standard deviation is the square root of the variance, and:
$$\begin{array}{rcl}
\mathrm{Var}(X)&=&\displaystyle \mathrm{E}\{X^2\}-\mathrm{E}^2\{X\}\\
&=&\displaystyle \sum_i (p_ix_i^2)-\mu^2\\
\end{array}$$
Where
$$\mu=\mathrm{E}\{X\}=\sum p_ix_i=p_1+2p_2+\ldots+dp_d$$
Therefore
$$\mu^2=\mathrm{E}^2\{X\}=\bigl(\sum_i p_ix_i\bigr)^2=\sum_i\sum_j p_ix_ip_jx_j=(p_1+2p_2+\ldots+dp_d)^2$$
And
$$\begin{array}{rcl}
\mathrm{Var}(X)&=&\displaystyle \sum_i (p_ix_i^2)-\mu^2\\
&=&\displaystyle \sum_i \bigl(p_ix_i^2\bigr)-\sum_i\sum_jp_ix_ip_jx_j\\
&=&\displaystyle \sum_i \bigl(p_ix_i^2\bigr)-\biggl(\sum_i\sum_j p_ix_ip_jx_j\biggr)\\
&=&\displaystyle \sum_i \bigl(p_ix_i^2\bigr)-\biggl(\sum_i p_i^2x_i^2+\sum_i\sum_{j,j\neq i} p_ix_ip_jx_j\biggr)\\
&=&\displaystyle \sum_i \bigl(p_ix_i^2\bigr)-\sum_i p_i^2x_i^2-\sum_i\sum_{j,j\neq i} p_ix_ip_jx_j\\
&=&\displaystyle \sum_i \bigl(p_ix_i^2-p_i^2x_i^2\bigr)-\sum_i\sum_{j,\,j\neq i} p_ix_ip_jx_j\\
&=&\displaystyle \sum_i p_i(1-p_i)x_i^2-\sum_i\sum_{j,\,j\neq i} p_ix_ip_jx_j\\
\end{array}$$
