Was hoping someone could just provide a logic check of my work, as some of my results did not match up to answers I saw online.
Let $X_{n}$ denote the maximum of n independent dice rolls. Let's first find the CDF:
$P(X_{n}\leq x) = \prod P({X_{i}} \leq x) = \left (\frac{x}{6} \right )^{n}$
Now we can take the derivative to find the PDF:
$= n * \frac{1}{6} * \left (\frac{x}{6} \right )^{^{n-1}}$
Now we can integrate to find the expected value:
$ =\int_{1}^{6} x*n * \frac{1}{6} * x^{n-1} * \frac{1}{6}^{n-1} dx =\int_{1}^{6} n * x^{n} * \frac{1}{6}^{n} dx = \frac{n}{n+1} * \left (\frac{1}{6} \right )^{n} * x^{n+1} $
And then we evaluate the definite integral at x = 6 - x = 1 (sorry I'm really bad with LateX)
Does this look right?
This is correct:
Let $X_{n}$ denote the maximum of n independent dice rolls. Let's first find the CDF: $$P(X_{n}\leq x) = \prod P({X_{i}} \leq x) = \left (\frac{x}{6} \right )^{n}.$$ Now the probability mass function (not a density as $x$ can only take discrete values $1, \ldots, 6$) is $$P(X_{n}= x) = P(X_{n}\leq x)- P(X_{n}\leq x-1)= \left (\frac{x}{6} \right )^{n}-\left (\frac{x-1}{6} \right )^{n}.$$ Then $$E[X_{n}] = \sum_{x=1}^6 xP(X_{n}= x) = \sum_{x=1}^6 x\left(\left (\frac{x}{6} \right )^{n}-\left (\frac{x-1}{6} \right )^{n}\right).$$ This is, collecting together terms in $\left(\dfrac{r}{6}\right)^n$ for different $r$, $$6-\sum_{r=1}^5\left(\dfrac{r}{6}\right)^n.$$ The sum doesn't have a very nice closed form, though for large values of $n$, $\sum_{r=1}^5\left(\dfrac{r}{6}\right)^n \approx\left(\frac{5}{6}\right)^n$ (relative error of $10^{-2}$ for $n=20$; $2 \times 10^{-10}$ for $n=100$).