Tail Sum Formula states that:
For $X$ with possible values $\{0, 1, 2, \ldots , n\}$, $$\operatorname E(X) = \sum_{j=1}^n P(X \ge j)$$
Suppose that 4 dice are rolled. Find the expected maximum $\operatorname E(M)$ of the 4 rolls.
$M$ has possible values $\{1, 2, \ldots, 6\}$ all consecutive. Thus, we can use the Tail Sum Formula.
\begin{align} \operatorname E(M) & = \sum_{j=1}^{6}P(M \ge j) \\[10pt] & = P(M \ge 1) + P(M \ge 2) + P(M \ge 3) + P(M \ge 4) + P(M \ge 5) + P(M \ge 6) \\[10pt] & = \left(\frac 6 6 \right)^4 + \left(\frac 5 6 \right)^4 + \left(\frac 4 6 \right)^4 + \left(\frac 3 6 \right)^4 + \left(\frac 2 6 \right)^4 + \left(\frac 1 6 \right)^4 \\[10pt] & = 1.755 \end{align}
But the expected minimum is also $1.755$!
Edit 1:
Let $X_i$ be the value obtained on roll $i$, $1 \le i \le 4$.
$P(M \ge j) = $ Probability that the maxium of these 4 rolls is $\ge$ j
$= 1 - P(M \lt j)$
$= 1 - P(M \le j-1)$ Probability that the maxium of these 4 rolls is $\le$ j
$= 1 - P(X_1 \le j - 1, X_2 \le j - 1, ..., X_6 \le j - 1)$
Since each roll is independent...
$= 1 - P(X_1 \le j - 1) P(X_2 \le j - 1) ... P(X_6 \le j - 1)$
$= 1-(\frac{j-1}{6})^4$
\begin{align} \operatorname E(M) & = \sum_{j=1}^{6}P(M \ge j) \\[10pt] & = P(M \ge 1) + P(M \ge 2) + P(M \ge 3) + P(M \ge 4) + P(M \ge 5) + P(M \ge 6) \\[10pt] & = \left(1- (\frac 0 6)^4 \right) + \left(1- (\frac 1 6)^4 \right) + \left(1- (\frac 2 6)^4 \right) + \left(1- (\frac 3 6)^4 \right) + \left(1- (\frac 4 6)^4 \right) + \left(1- (\frac 5 6)^4 \right) \\[10pt] & = 5.244598765 \end{align}