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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?

Matthew Kaplan
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    Even easier: use the (easily demonstrated) fact that, for discrete variables supported in $\mathbb N$, we have $E[X]=\sum_{k=0}^{\infty} P(X>k)$. – lulu Apr 30 '23 at 16:58

1 Answers1

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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 $n=100$).

mcd
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  • The exact result in this answer is a closed form, though you might debate whether it is “nice.” It can also be written $6-(1+2^n+3^n+4^n+5^n)/6^n.$ – David K May 02 '23 at 01:14
  • Yes, I assume you mean the Bernoulli expression: certainly not very nice. – mcd May 02 '23 at 05:27
  • No, I meant exactly what you wrote, because (unlike the Faulhaber formulas) you have exactly the same number of terms in the sum (five terms) no matter what number of rolls are involved. – David K May 02 '23 at 05:29
  • I see what you mean - and I agree; a 5 term term sum is certainly a closed form, and not even that wearisome to compute. – mcd May 02 '23 at 13:39