What are the variance and standard deviation for the bin counts for n rolls on a standard six-sided die? I'm having trouble imagining what variance and deviation mean with a series of die rolls.  That is, a fair die will fall with a flat distribution on all its values 1-6 in 6 bins (1, 2, 3, 4, 5, 6) over time (as n goes towards infinity).
Firstly, does the concept of variance really make sense on such a question? [Edit: only if I provide some data on bin outcomes.  Say n=36, and the die lands as follows:  1 (6 times), 2 (5x), 3 (5x), 4 (7x), 5 (7x), 6 (6x).]
The average outcome will be n/6 over time for each of the six bins  [Edit:  My prior writeup was confusing, as I had said the mean was 3.5 -- but this mean face-value is irrelevant to the question.]
Is this question even valid?  It seems a perfectly flat distribution (as n-> infinity), with no other hidden variables, has no variance (or shouldn't have any), but then what should one make of the results when n is finite?
 A: If $X$ is the value of the die we already know $\text{E}(X) = 21 / 6$ so we only need to find $\text{E}(X^2)$ since $\text{Var}(X) = \text{E}(X^2) - \text{E}(X)^2$.  We can just directly calculate
\begin{align}
\text{E}(X^2) &= \sum_{k=1}^{6} \frac{k^2}{6} \\
&= \frac{1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2}{6} \\
&= \frac{91}{6}
\end{align}
which after some arithmetic gives us $\text{Var}(X) = 105 / 36$.
A: This is a discrete uniform distribution.  So we can use $\frac{(b-a+1)^2-1}{12}$ to solve for the variance. $\frac{(6-1+1)^2-1}{12}$ = $\frac{6^2-1}{12}$ = $\frac{35}{12}$
A: There are already several good answers posted (as well as one in the comments). My goal here is not to replicate those answers, but rather to try and address an apparent confusion about the "definition of variance".
In your question you say

It seems the variance and standard deviation tacitly ASSUME an a priori normal distribution around an unspecified or unknown order -- but a flat "curve" with no other hidden variables has no variance.

And in the answer you posted, you say

The answer should be (ahem: is) 0. Apparently the equations for variance assume another unknown variable (another dimension) affecting results.

If we call the value of a die roll $x$, then the random variable $x$ will have a discrete uniform distribution. That is, if we denote the probability mass function (PMF) of $x$ by $p[k]\equiv\Pr[x=k]$, then we have $p[k]=\frac{1}{K}$, where $K$ is the number of distinct values $k$ can take (i.e. here $K=6$).
Independent of the form of the probability distribution, the mean $\mu$ and variance $\sigma^2$ are always defined in terms of expectations. These definitions are
$$
\mu_x\equiv\mathbb{E}[x] \,,\, \sigma^2_x\equiv\mathbb{E}\left[(x-\mu_x)^2\right]
$$
(e.g. see Wikipedia).
For a discrete random variable such as $x\in\{X_1,\ldots,X_K\}$ with PMF $p[X_k]\equiv\Pr[x=X_k]$, the expectation operator $\mathbb{E}[\,]$ is defined by
$$
\mathbb{E}\big[f[x]\big]\equiv\sum_{k=1}^Kf[X_k]p[X_k]
$$
where $f[\,]$ is any deterministic function.
Your confusion appears to be related to this last part. For the mean $\mu$ you appear to be correctly using $f[x]=x$. However, for the variance you appear to be using $f[x]=p[x]$, i.e. the PMF of $x$.
Perhaps the following summary will make things more clear
\begin{array} {c|c|c}
\text{object }(f) & \text{mean }(\mu_f) & \text{variance }(\sigma_f^2) \\
\hline
x & \frac{7}{2} & \frac{105}{36} \\
p[x] = \frac{1}{6} & \frac{1}{6} & 0
\end{array}
In other words, the probability distribution $p[x]$ has zero variance, but the die value $x$ certainly has non-zero variance.
A: While @dsaxton's answer is correct, I think it makes it more difficult for beginners in statistics to grasp the concept of variance, so I'll offer another answer that helps you get a better "feel" for the what the variance is actually "doing."  An equivalent expression for the variance in this case is:
$Var(X)$ =$ \sum_{i=1}^6(X_i-\bar{X})^2\over{6}$.  
Now, you know the mean, $\bar{X}=3.5$, so you simply need to take the die's $i$th's face value $i=1, 2, . . . , 6$, $X_i$ and subtract it from the mean, square it, and divide it by 6.  In effect this gives you an average of how far away each die value is from its mean.  So $Var(X)$ is given by:
${(1-3.5)^2+(2-3.5)^2+(3-3.5)^2+(4-3.5)^2+(5-3.5)^2+(6-3.5)^2}\over{6}$= $17.5\over{6}$=$105/36$, the same answer @dsaxton provided.  
We square the values of $X_i-\bar{X}$ because if we don't, then the sum of the values will add to zero and the negative numbers will cancel out the positive numbers.
A: "That doesn't make sense: if the die starts at zero-based numbering, then the variance changes. But, variance shouldn't depend on the numbering of the die as all faces are equally probable. –
Marcos
Dec 7, 2016 at 20:38"
It seems to me the variance is not the 'variance of the probability' of one side over another, but is the 'variance of the face values' from the throws.
