Expected Value of Dice Prospect I am facing a problem where I have $n$ 6-sided dice that I roll all at once,
and I earn the minimum dollar face of the dice rolls. For example, if I have 3
6-sided dice, and I roll $1,2,$ and $4,$ I earn $\$ 1$. 
I want to find the expected value of these dice rolls. The support seems rather
straightforward, and it seems to me that the case where the outcome is $6$ is
rather trivial (since there is only one way that this can occur). However, I
am running into several double-counting issues when I attempt to get the
probability of the other support values. I was wondering if someone could give
me suggestions on how I go about those counting arguments.
 A: Let $M$ denote the minimum of $n$ 6-sixed (presumably fair) dice, and let $X_i$ be the value of the $i$th die. Then
\begin{align*}
P(M\geq m) &= P(X_1 \geq m, \dotsc,X_n\geq m)\\
&= P(X_1 \geq m)\dotsb P(X_n \geq m) \tag 1 \\
&=\left(\frac{6-m+1}{6}\right)^n
\end{align*}
where $(1)$ is true by independence.
For the expectation, you might be tempted to find $P(M=m)$, however, recall that you can find the expectation by tail sum
\begin{align*}
E[M] &=\sum_{m = 1}^6 P(M\geq m)\\
&=\frac{1}{6^n}\sum_{m=1}^6 (7-m)^n\\
&=\frac{1}{6^n}\left[6^n+5^n+\dotsb+1^n\right]\\
&=\frac{1}{6^n}\sum_{k = 1}^6k^n.
\end{align*}
A: Take the case $n=3$. Imagine the cube of possible outcomes. $1$ has an outcome of at least $\$6$, $8$ will pay at least $\$5$, $27$ will pay at least $\$4$, etc...
In the case case $k^n$ will pay at least $\$7-k$.
A: We will use an alternate formula for expectation that applies to non-negative integer-valued random variables. For a proof, please see this, near the end.
$$E(X)=\sum_{i=1}^\infty \Pr(X\ge i).$$
In our particular case, with $n$ rolls, we have $\Pr(X\ge i)=\frac{(7-i)^n}{6^n}$ for $i=1$ to $6$. Thus
$$E(X)=\sum_{i=1}^6 \frac{(7-i)^n}{6^n}.$$
