That’s because we aren’t trying to look at one specific uniform distribution. We were asking why Benford’s law happens for almost all processes that follow a uniform distribution and record the result as positional notation with digits — namely that 1 appears a lot more than 2, which appears a lot more than 3, etc. Roughly in the proportion that 1 is twice that of 2, which is 1/3 more than 3, etc.

(Btw it is NOT true for eg dictionary words for example, an initial A doesnt appear more than B. That should tell you something!)

And to understand the reason we just have to look at the family of uniform distributions, and see that for almost all of them, this proportion holds. Sure, for some of them, the 1,2,3 may be even MORE prevalent relative to 4-9 because the maximum value was 400 or 4000 or 40000. Ok? You can see this. For a uniformly distributed process that happens to have that as the maximum, Benford’s law will have the same proportions between 1,2,3 but then drop for 4-9 since they didn’t get that “boost”.

But if you keep sampling and this maximum keeps growing by some continuous distribution that’s not perfectly synced with the metric system, then it’s as likely to be in the range 100-200 as it is to be in 200-300. And then as likely to be in 1000-2000 as in 2000-3000. Given that, we get something like Benford’s law.

Now, perhaps it is ALSO TRUE for other distributions. I just explained why it’s true for uniform ones.

If you took a random letter in the alphabet and then sampled from any letter before this letter you would get more samples from the earlier letters of the alphabet. That is because the two step process discards higher letters in the first step. This is not a uniform distribution and is not Bensford's law. It's just a weird two step process that over-samples earlier letters.

You are right, it would — but that’s not how it works. People don’t sample words by beginning with AAA and moving on to larger ones. So your point is just being put out there to “win”?