Rolling dice against each other? After doing some research, I found that the probability of rolling an n-sided die against another, equivalent die and landing a higher number on the first rolled dice to be ((n−1)/2)/n, where n is the number of faces.
How do I calculate this probability for multiple dice rolled against each other?
 A: I'll start you off.
Let's say you have $M$ $N$-sided dice that are white, and one $N$-sided die that is green.  You roll them all, and the green one shows a value of $R$.  (The faces are labeled $1 ... N$.)
In order for the green die to have the largest value, all of the others must have a value less than $R$.  There are $R-1$ such values out of $N$, so the number of "winning" combinations of values is $(R-1)^M$.  There are $N^M$ total combinations for the white dice, and so the probability of the green die being the highest, given that it shows a value of $R$, is
$$P(R) = \left[\frac{R-1}{N}\right]^M.$$
The solution to your question involves considering all values of $R$.  Can you take it from here?  (Note, the answer you get should agree with the one you already have by setting $M=1$.)
A: Suppose you have $k$ identical fair dice, each of which $n$-sided with all sides different (without loss of generality, labeled $1,2,3,\dots,n$ though this is unimportant).
Suppose that we roll each die in sequence independently and we ask what is the probability that the first die rolled is in fact strictly larger than any other seen result of the remaining dice.
We can approach via multiplication principle and addition principle.
In the event that the result of the first die is $i\in \{1,2,\dots,n\}$, for each other die rolled the probability that it is in fact strictly lower than $x$ will be $\frac{i-1}{n}$.  The probability that all dice after the first are lower then is $\left(\frac{i-1}{n}\right)^{k-1}$.  The probability that the first die shows $i$ in the first place will be $\frac{1}{n}$
Ranging over all possible values of $i$ then, we have the following summation:
$$\sum\limits_{i=1}^{n}\left(\frac{i-1}{n}\right)^{k-1}\cdot\frac{1}{n}$$
checking to make sure the formula makes sense:
In the case $k=1$, every term in the summation is equal to $\frac{1}{n}$ and the total summation is equal to $1$.  This is what we expected because if we only roll one die it will clearly be the largest result.
In the case $k=2$, we have the $(n-1)^\text{st}$ triangle number as a numerator, simplifying as $n(n-1)/2$, and $n^2$ as a denominator, yielding the probability as $(n-1)/(2n)$ as expected.
The above summation might simplify using Generalized Harmonic Numbers if you so choose, but it may be simplest left as is.
