<s>[INST] What is the average of rolling two dice and only taking the value of the lower dice roll? My question is related to this question, with the exception that I'm looking for the average when taking only the lower of two dice rolls.
My question is:

What is the average of rolling two dice and only taking the value of the lower dice roll?

This formula is used for the "take higher roll" case:
$$
E[X] = \sum_{x=1}^6\frac{2(x-1)+1}{36}x = \frac{1}{36}\sum_{x=1}^6(2x^2 - x) = \frac{161}{36} \approx 4.47
$$
...and includes the term "$2(x−1)+1$". Could someone please explain how that term would have to be changed so that it applies for "take lower roll"? From the original example, I'm somewhat confused where "$x-1$" comes from, although it is mentioned in the comments.
PS: I would have written this question as a comment to the original question, but I lack the points to be allowed to write comments.
 [/INST] Let the random variable $X$ be defined as the maximum value when two fair six-sided dice are rolled.  We wish to find $E[X]$.
Since there are $k$ possible values which are at most $k$, there are $k^2$ ordered pairs of values in which both dice display a number which is at most $k$.  To ensure that the maximum value is $k$, we must subtract the $(k - 1)^2$ cases in which both values are at most $k - 1$.  Since there are $6^2 = 36$ possible ordered pairs of values, the probability that the maximum value is $k$ is
$$\Pr(\max = k) = \frac{k^2 - (k - 1)^2}{36} = \frac{k^2 - (k^2 - 2k + 1)}{36} = \frac{2k - 1}{36}$$
Since the maximum value can assume any integer value from $1$ to $6$, the expected value for the maximum value is
\begin{align*}
E[X] & = \sum_{k = 1}^{6} k\Pr(X = k)\\
     & = \sum_{k = 1}^{6} \frac{k(2k - 1)}{36}\\
     & = \sum_{k = 1}^{6} \frac{2k^2 - k}{36}\\
     & = \frac{1}{18} \sum_{k = 1}^{6} k^2 - \frac{1}{36}\sum_{k = 1}^{6} k\\
\end{align*}
Use the formulas
\begin{align*}
\sum_{k = 1}^{n} k^2 & = \frac{n(n + 1)(2n + 1)}{6}\\
\sum_{k = 1}^{n} k & = \frac{n(n + 1)}{2}
\end{align*}
with $n = 6$ to finish the calculation.
Let the random variable $Y$ be defined as the minimum value when two fair six-sided dice are rolled.  We wish to find $E[Y]$.
On a six-sided die, there are $7 - k$ values which are at least $k$. Hence, there are $(7 - k)^2$ ordered pairs of values in which both dice display a number which is at least $k$.  To ensure that the minimum value is $k$, we must subtract the $[7 - (k + 1)]^2 = (6 - k)^2$ cases in which both dice display a number which is at least $k + 1$.  Since there are $6^2 = 36$ possible ordered pairs of values, the probability that the minimum value is $k$ is
$$\Pr(\min = k) = \frac{(7 - k)^2 - (6 - k)^2}{36} = \frac{49 - 14k + k^2 - (36 - 12k + k^2)}{36} = \frac{13 - 2k}{36}$$
Since the minimum value can assume any integer value from $1$ to $6$, the expected value for the minimum value is
$$E[Y] = \sum_{k = 1}^{6} k\Pr(Y = k) = \sum_{k = 1}^{6} \frac{k(13 - 2k)}{36}$$
I will leave it to you to finish the calculations.
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