When rolling multiple dice, what is the average value if you only keep the highest roll? These questions come up regarding a RPG in which you roll multiple dice, but only keep the highest value. So, suppose I were to roll 5 dice, each of which has 12 sides (numbered from 1 to 12). If I only keep the highest roll, what would the average end up being?
To throw in an additional level of complication, what would happen if the dice "explode" or "penetrate"?
Explode/Penetrate=If you roll a die and roll the highest value (if you roll an 8 on an eight sided die), re-roll the die and add the results together. Repeat until you do not roll the maximum value.
 A: Here is the answer for the first part of your question:
There are one event, where the maximal number is $1$ (all dices showing $1$). 
There are $2^5$ events where all dices showing either $1$ or $2$. Subtract the one event, where all dices showing $1$, you get $2^5-1$ events, where maximal number is $2$.
There are $3^5$ events, where all dices showing $1$, $2$ or $3$. Subtract the $2^5$, where the maximal number is $1$ or $2$ and you get $3^5-2^5$ events with maximal number $3$.
Continue that idea and you get the average:
$\frac{1}{12^5}\sum\limits_{i=1}^{12}(i^5-(i-1)^5)\cdot i $
(Note that $12^5$ is the whole number of possible events)
A: For your first question (expected value for max value of 5 rolls of a 12-sided die):
Let $X$ be the maximum value.
Then $P(X=k)=\left(\frac{k}{12}\right)^5-\left(\frac{k-1}{12}\right)^5$ for $k=1,2,...,12$.
The reasoning is that for the maximum to be $k$, you need all the rolls to be less than or equal to $k$, but not all less than $k$.
So $$E[X]=\sum_{k=1}^{12}k \left( \left(\frac{k}{12}\right)^5-\left(\frac{k-1}{12}\right)^5\right)=\frac{2604108}{12^5}\approx10.4653$$
