Probability of the maximum of a throw of an infinite number of $n$-sided dice being $k$ [closed]

Question

Let $X$ be the random variable obtained as the maximum of a throw of $m$ dice (each of which is $n$-sided). In other words, $X = \max\{l_1,\cdots, l_m\}$ where $l_i$ can take any value between $1$ and $n$. Using the relationship between the sum of consecutive powers and Bernoulli numbers, one obtains the following formula for the probability of $X=k$:

$$Pr(X=k) = \frac{1}{n^m}[k^m -\frac{1}{m}B_m(k) + \frac{1}{m}B_m)]$$

where $B_m$ is the $m$th Bernoulli number and $B_m(k)$ is the $m$th Bernoulli polynomial evaluated at $k$. Clearly, if we let $n$ approach infinity, then the quantity above will approach zero. However, I am unsure what happens if we instead let $m$ approach infinity. The question then is does the following limit exist

$$Pr_{\infty}(X=k) := \lim_{m\to\infty}(\frac{1}{n^m}[k^m -\frac{1}{m}B_m(k) + \frac{1}{m}B_m)]) $$

Answer 1

Intuition should say that if we keep tossing a die with a finite number of outcomes, then sooner or later we will see every outcome, in particular, we will see the maximum outcome. A proof is

$$\Pr[\text{no throw equals $n$}] = \Pr[l_1\neq n \text{ and } \dots \text{ and }l_m\neq n]$$ $$ = \Pi_{j=1}^m \left(1-\frac{1}{n}\right) $$ $$ = \left(1-\frac{1}{n}\right)^m $$ $$ \to 0 \text{ as } m \to \infty .$$

So with probability approaching $1$, some throw equals $n$ and the maximum equals $n$ (so the probability that the max is $k < n$ goes to zero). I guess this doesn't say that the limit of your expression exists, but if it does, it's zero for $k < n$ and one for $k=n$.

(Edit: If you are interested in getting more comfortable with arguing this sort of thing, I like Mitzenmacher and Upfal's Probability and Computing, which is mostly about probability tools with applications to computing -- but then I'm in CS. Anyway, for finite cases, this sort of problem might be called "balls into bins" or "coupon collectors".)