Trouble finding var(ax) So the variance of a 6-sided (1,2,3,4,5,6) die is $291.6$ using the formula:
$$
\text{Var}(X) = \frac{(b-a+1)^2}{12}
$$
Also, $\text{Var}(10X) = 10^2 \cdot \text{Var}(X)$, so that would mean $\text{Var}(10X) = 291.6$.
If I want to find the variance of $10X$, is this not the same as multiplying each of my values by $10$? So that I'm now finding the variance of $(10,20,30,40,50,60)$? If I use the same formula as above, my answer is $216.67$, which is not equal to $291.6$ which I expected. Not exactly sure which part of my logic is incorrect.
 A: Let's find a formula that will apply to both your situations.
One description that covers them both supposes $X$ is a uniform random variable defined on an arithmetic progression
$$x_1, x_2, \ldots, x_n = a, a+d, a+2d,\ldots, a+(n-1)d = b.$$
Thus $x_i=a+(i-1)d$ and each $x_i$ has a probability $1/n.$  By definition
$$E[X] = \sum_{i=1}^n \Pr(x_i)x_i = \sum_{i=1}^n \frac{1}{n}\, (a+(i-1)d) = \frac{2a+(n-1)d}{2} = \frac{a+b}{2}.$$
Then, also by definition,

$$\begin{aligned}
\operatorname{Var}(X) &= E[(X-E[X])^2] = \sum_{i=1}^n \Pr(x_i)(x_i-E[X])^2 \\
&= \sum_{i=1}^n \frac{1}{n}\, \left(a + (i-1)d - \frac{2a+(n-1)d}{2}\right)^2\\
&=d^2\frac{n^2-1}{12}.
\end{aligned}$$

The factor of $d^2$ is precisely what you expected from the scaling law for the variance.  Let's check.

*

*The standard die is described by $a=1,$ $d=1,$ and $n=6,$ so that $$d^2\frac{n^2-1}{12} = (1)^2 \frac{6^2-1}{12} = \frac{35}{12} = 2.91\bar6.$$


*Upon multiplying by $10$ we have $a=10,$ $d=10,$ and $n=6$ still, so that $$d^2\frac{n^2-1}{12} = (10)^2 \frac{6^2-1}{12} = 100 \frac{35}{12} = 291.\bar6.$$
When you obtained $216.67,$ you were applying the formula $((b-a+1)^2 - 1)/12$  (notice the additional "-1" in the numerator).  But in terms of $a,$ $n,$ and $d,$ this is
$$\frac{(b-a+1)^2-1}{12} = \frac{(a+(n-1)d - a + 1)^2-1}{12} = \frac{(d(n-1)+1)^2}{12}$$
which gives the correct value only when $d=0$ or $d=1.$ Your formula does not apply to any other situation.  That's why we needed to work out the generalization.
Finally, if you would prefer a formula in terms of the two endpoints $a$ and $b$ and the count $n\gt 1,$ you can recover $d$ as
$$d = \frac{b-a}{n-1}$$
and plug that in to get

$$\operatorname{Var}(X) = \left(\frac{b-a}{n-1}\right)^2 \frac{n^2-1}{12} = \frac{(b-a)^2}{12}\,\frac{n^2-1}{(n-1)^2}.$$

This is informative, because for medium to large $n,$ the second fraction is close to $1$ (the error is on the order of $1/n$) and can be ignored.  What is left is the variance of the variable uniformly distributed over all numbers between $a$ and $b.$  The quadratic dependence on the scale is explicit in the factor $(b-a)^2.$
