Variance for the minimum of rolling two six-sided dice. What is the variance for the random variable which is the minimum value of rolling two six-sided dice?
I don't know how to approach this problem. To get the variance, I thought I'd calculate the expected value but to do that, I need to use total expectation, and there will be 36 or so different partitions.
 A: For any nonnegative integer valued random variable, $\mathsf E(Y)=\sum_{k=1}^\infty \mathsf P(Y\geq k)$
$$\begin{align}\mathsf E(\min\{X_1,X_2\})~&=~\sum_{k=1}^6 \mathsf P(\min\{X_1,X_2\}\geq k)\\ &= \sum_{k=1}^6 \mathsf P(X_1\geq k, X_2\geq k)\\&=\tfrac 1{36}\sum_{k=1}^6(7-k)^2 \\ &= \tfrac 1{36}\sum_{j=1}^6 j^2
\end{align}$$
Similarly: $\mathsf {E}(Y^2) =\sum_{k=1}^\infty (2k-1)\,\mathsf P(Y\geq k)$ , so:
$$\mathsf E(\min\{X_1,X_2\}^2)=\tfrac {15}{36}\sum_{j=1}^6 j^2-\tfrac 2{36}\sum_{j=1}^6 j^3$$
A: Yes, there are a lot of options; but, we can reduce them through a bit of cleverness.
Instead of coming up with the mass function, let's come up with a CDF; since the outcome is discrete, we can easily compute the mass function from there.
To that end: note that for $i\in\{1,2,\ldots,6\}$, if the first and second rolls are $X_1$ and $X_2$, then
$$
P(\min\{X_1,X_2\}\leq i)=1-P(\min\{X_1,X_2\}>i)=1-P(X_1>i,X_2>i)=1-P(X_1>i)P(X_2>i).
$$
These should, of course, be easily computed. Now, notice that
$$
P(\min\{X_1,X_2\}=i)=P(\min\{X_1,X_2\}\leq i)-P(\min\{X_1,X_2\}\leq i-1),
$$
giving you a PMF.
A: Problems like this can be easily solved using order statistics. In particular, we are given a parent random variable $X \sim \text{DiscreteUniform}(1,6)$ with pmf say $f(x)$:

The minimum of two iid samples drawn from the parent corresponds to the $1^\text{st}$ Order Statistic (i.e. sample minimum) in a sample of size 2, which has pmf say $g(x)$:

for $x \in \{1, 2, \dots, 6\}$, and where I am using the OrderStat function from the mathStatica package for Mathematica to automate the calculation.
 Here is a plot of the pmf of the minimum of two dice $g(x)$:

We seek $\text{Var}(X)$ when $X$ has pmf $g(x)$:

which is approximately 1.97145. 
Comparison to other solutions
The second moment $E_g[X^2]$ is:

which is different to the solution posted by Graham Kemp of $\mathsf E(\min\{X_1,X_2\}^2)=\tfrac {15}{36}\sum_{j=1}^6 j^2-\tfrac 2{36}\sum_{j=1}^6 j^3$ which evaluates to $\frac{161}{12}$ ( referred to @GrahamKemp )
Notes
As disclosure, I should add that I am one of the authors of the Var and OrderStat functions used above.
