Dice games, part 2

I apologize if this is unnecessarily confusing or complicated - I'm not well versed in probability theory (or calculus, for that matter). This is just the way I solved it.

There are $k^j$ ways to roll only $k$ 's or less when rolling $j$ dice.

Therefore there are $k^{j}-(k-1)^{j}$ ways for $k$ to be the greatest roll.

The expected value of rolling $j$ $n$ -sided dice is therefore

$E_{n,j}=\frac{\sum_{k=1}^{n}k(k^{j}-(k-1)^{j})}{n^{j}} = n^{-j} \sum_{k=1}^{n} \left[k^{j+1}-k(k-1)^{j}\right]$

By the Binomial Theorem:

$=n^{-j}\sum_{k=1}^{n}\left[k^{j+1}-k^{j+1}+jk^{j}-{j\choose2}k^{j-1}+{j\choose3}k^{j-2}-\ldots \pm k\right]$

$=n^{-j}\left[j\sum_{k=1}^{n}(k^{j}) - {j\choose2}\sum_{k=1}^{n}(k^{j-1}) + {j\choose3}\sum_{k=1}^{n}(k^{j-2}) - \ldots \pm \sum_{k=1}^{n}k\right]$

$=n^{-j}\sum_{t=1}^{j}(-1)^{t+1} {j \choose t}\sum_{k=1}^{n}k^{j-t+1}$

We are asked to find

$\lim_{n \to \infty} \frac{E_{n,20}}{E_{n,30}} = \lim_{n \to \infty} \frac{n^{-20}\sum_{t=1}^{20}(-1)^{t+1} {20 \choose t}\sum_{k=1}^{n}k^{21-t}}{n^{-30}\sum_{t=1}^{30}(-1)^{t+1} {30 \choose t}\sum_{k=1}^{n}k^{31-t}}$

The $t=1$ term of the outer sums are the only terms relevant to the limit.

$= \lim_{n \to \infty} n^{10}\frac{20\sum_{k=1}^{n}k^{20}}{30\sum_{k=1}^{n}k^{30}}$

I will demonstrate that $\sum_{q=1}^{\infty}q^p$ is asymptotically equivalent to $\lim_{n \to \infty}\frac{n^{p+1}}{p+1}$ .

$\int_{0}^{m}k^{p}dk=\frac{m^{p+1}}{p+1}$

$=\lim_{n\to\infty}\sum_{q=1}^{n}\left(\frac{mq}{n}\right)^{p}\left(\frac{m}{n}\right)=\lim_{n\to\infty}\sum_{q=1}^{n}\frac{m^{p}q^{p}}{n^{p}}\left(\frac{m}{n}\right)$

$=\lim_{n\to\infty}\frac{m^{p+1}}{n^{p+1}}\sum_{q=1}^{n}q^{p}=\frac{m^{p+1}}{p+1}$

$\lim_{n\to\infty}\frac{1}{n^{p+1}}\sum_{q=1}^{n}q^{p}=\frac{1}{p+1}$

implying that $\sum_{q=1}^{\infty}q^p \sim \lim_{n \to \infty}\frac{n^{p+1}}{p+1}$ .

Substituting:

$\lim_{n \to \infty} n^{10}\frac{20\sum_{k=1}^{n}k^{20}}{30\sum_{k=1}^{n}k^{30}} = \lim_{n \to \infty} n^{10}\frac{20\frac{n^{21}}{21}}{30\frac{n^{31}}{31}} = \frac{20*31}{30*21} = \frac{62}{63}$

$62 + 63 = \boxed{125}$

Same solution here I used on Dice game, except this time it actually works

For a number randomly selected between a range (If there are infinite numbers in between, for some reason it breaks down if not), the average maximum is going to be 1/2 the maximum of the range. This is common sense, as it has a equal chance of being on either side. For two picks, the average maximum would be 2/3rds. Any pick would have a equal chance of being in each of the three thirds, so it will average at the middle of the thirds. And so on, such that randomly picking a number p times would give an average max of p/p+1.

So, the top number is 20/21 and bottom is 30/31.

20*31 = 620

30*21 = 630

620/630 = 62/63

62 + 63 = 125

My scientific method appeared to me with 0.98414 and Excel gives a nearest fraction of $\frac{62}{63}$ to it. I checked and divided, 0.984126984126984 comes to me. At the moment I see the screen appeared with 0.98412..., I realized that my initial thought of $\frac{49207}{50000}$ is unnecessary, because 62 + 63 = 125 shall be a very likely answer wanted comparatively, in a context of relating to figures of 20 and 30.

The idea is logical and I interpreted as making a further dicing after the 20th to continue up to 30th must make a higher or equal maximum with 100% certainty. I checked my proper simulation of command for correct call for $Randomize$ to observe for good random numbers generated. Such positive figure of less than 1 is actually resembling such a dice described of which I thought very suitable to the question.

True! Just apply computing technique to record for two maximums for an unfixed ratio and run for a dynamic average. The dynamic average is what we can see a constant to happen in this dicing!

$\frac{63}{64}.$ I credited this to excellent random generator. Likeliness is crucial to my guessing.

Answer: $\boxed{125}$