Have I found something new? (a formula for the odds of rolling at least one 1 when rolling an n-sided die n times)

Question

The formula

First, the formula I found for the odds of rolling at least one 1 when rolling an n-sided die n times:

Math.round(1 - ( ((n-1)/n)^n -1 )*(n^n-1) )/n^n

AKA:

Note: As n grows, the Math.round becomes less necessary.

Now, here's how I got there:

I started with the basics:

I then continued with the basics, but noticed some patterns:

In particular, I noticed that the number of outcomes that included at least one 1 could be determined from a sum that is based on the number of first 1's in a row, per column (note that the first 1's in each row are bolded in the above image, their counts shown at the bottom of each column, and that the non-first 1's in each row are made red):

I then recast the sum as a definite integral:

From there I used Wolfram to crank out an estimated solution for the number of outcomes that include at least one 1 (some hand-waving steps have been left out...because I don't remember them):

And bing-bang-boom, as long as I round that estimated number of outcomes containing at least one 1 before dividing it by the total number of outcomes (n^n), I got the right answer!

In closing

I don't think that this formula is better for computational solutions for the odds of getting at least one 1 when rolling an n-sided die n times -- because of the fast growing n^n... buuut, I do feel like this solution exposes some sort of potentially unique connection between the physical world and the realm of probabilities... admittedly, I'm not good at statistics and recognize that my googling skills may also have failed me when trying to find a similar analysis.

PS:

The leap I made to using summation notation came from my playing around with methods for representing the odds for different values of n using MS Excel. Please note that the appearance of "!" in the final image below denotes my joy at finding answers to the questions I'd posed in the first two of these final three images, not factorials.

Answer 1

I didn't need to attempt to work out the integral version of the summation I'd found that represented the number of times at least one 1 would appear when rolling an n-sided die n times.

Instead, there is a direct answer, right in one of the tables I'd shown, for what my summation becomes.

Specifically, I noticed that there was a pattern to the differences between the numerator and denominator values that represented the odds for each listed n. Following up on it this time, I noticed that the difference was $(n-1)^n$.

Thus, the numerator would be the denominator minus the difference, aka:

$$n^n - (n-1)^n$$

Which, when divided by the denominator ($n^n$) yields the familiar formula for the situation at hand.

(side note, Wolfram Alpha verified for me that the summation becomes $n^n - (n-1)^n$)

The summation:

Becomes the numerator in the following fraction:

And the fraction simplifies to the most common form of the answer:

Now I can sleep.