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

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


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  • $\begingroup$ Hint: $\mathbb P(l_i = n) = n^{-1} > 0$. $\endgroup$ – cardinal Oct 20 '14 at 0:42
  • $\begingroup$ Thanks for your hint. Yes, it is true, but the events $E_i$ given by $\{(l_j) \mid l_i = n\}$ are not mutually exclusive. I should say that neither probability theory nor analysis are my specialty so perhaps I missed the hint. $\endgroup$ – Andrew Stout Oct 20 '14 at 0:57

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".)

  • $\begingroup$ ah, yes, now the other hint makes sense too. Thanks for your help. Your answer implies that $(B_m(k)-B_m)/mn^m$ goes to zero as $m$ approaches infinity provided $n>1$. I thought that maybe there is a way to prove this second fact directly. $\endgroup$ – Andrew Stout Oct 20 '14 at 1:42
  • $\begingroup$ Actually, this way is very easy also $\sum_{j=0}^{k-1} j^m = \frac{1}{m}(B_m(k) - B_m)$ which will immediately imply the result which you proved. The Bernoulli numbers were a red-herring. $\endgroup$ – Andrew Stout Oct 20 '14 at 3:06

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