Is rolling 10% 10 times just as good as 1 100% or better? How do you determine the value of a percentage based on how many rolls? For example would 6 rolls at 10% be better than 1 roll at 60%? 1 roll at 100% to 100 rolls at 1%?
I know each probability is independent of each other, but I don't know how to determine which is a better odd.
 A: I assume you are asking which approach gives better odds of at least one successful outcome. 
Taking one roll with probability $p$ clearly succeeds with probability $p$.
Taking $n$ rolls each of which has probability $\frac{p}{n}$ has at least one success with probability
$$1-\left(1-\frac{p}{n}\right)^n$$
since it is one minus the probability of failure on each roll.
So in your first example, six rolls with success probability $\frac{1}{10}$ gives you a total success probability of
$$1-0.9^6\approx 0.469.$$
You can work out other examples yourself.
Note that
$$\lim_{n\to\infty}\left(1-\left(1-\frac{p}{n}\right)^n\right) = 1-e^{-p},$$
so that as the number of trials increases (and the success probability of each decreases) the probability of at least one success approaches $1-e^{-p}$. This is always less than $p$ unless $p=0$.
A: $n$ rolls at $p$ gives the probability
$1-(1-p)^n$
one roll at $np$ (assuming $np<1$)
the second one is always higher than the first one and they're equal for $n=1$
