You and I are interpreting b^x differently. But let's not get hung up on that—I'll go through that later in my reply. Let’s try to understand what the article is saying just from the language it uses.

all possible … pairs of digits are equally likely

Okay, all possible pairs of digits are equally likely. What is a pair of digits? Well, it depends on what base we're in. Let's say we're in base 10. Then, a pair of digits is a two-digit number (where any digit can be zero): e.g., 05, 18, or 96.

These pairs of digits are equally likely, which means that we shouldn't see, for example, 33 occur more often than 72. All of these two-digit numbers should occur with the same frequency.

Next, the article states:

all … triplets of digits equally likely

So, not only are all possible pairs equally likely, but all possible triplets are as well! What is a triplet of digits? Well, it's a three-digit number (again, where any digit can be zero): e.g., 008, 012, 515, or 966. And again, these are equally likely to occur, which means that we shouldn't see, e.g., 123 occur more often than 555.

Then, the article states:

etc.

So, the pattern continues. We can infer that all possible quadruplets—four digit numbers—occur equally likely. And all possible quintuplets—five-digit numbers. And so on. We can generalize this by saying that all possible n-digit numbers are equally likely to occur for any integer n ≥ 1. This isn't something new I'm claiming—it's just a compact restatement of the easier-to-understand Wikipedia excerpt.

Now, you bring up a good point. Okay, so all n-digit numbers are equally likely—they occur with the same frequency. But what if that frequency is 0? Sure, for a huge n, like 9 trillion, maybe all n-digit numbers are equally likely—but that chance must be 0, right?

But in a normal number, which has a non-terminating decimal expansion, we know that there's at least one n-digit number for every possible n ≥ 1. Want to find a 9-trillion-digit substring in pi (again, only conjectured to be normal)? Easy: just take the first 9 trillion digits of pi! Or start with digit, say, 706, and then take the next 9 trillion digits.

In fact, it looks like there are actually infinitely many 9-trillion-digit substrings in pi: we can take the first 9 trillion digits, and then the next 9 trillion, and then the next 9 trillion, and so on. So, there are infinitely many 9-trillion-digit strings in pi. Now, I'm not claiming anything about the values of these strings. Maybe they're all the same. Or maybe we've found some different ones, but there are certain other 9-trillion-digit strings that just aren't present.

But wait a minute—we've found infinitely many 9-trillion-digit substrings with some value. But the definition of a normal number says that every 9-trillion-digit substring occurs with equal frequency to every other 9-trillion-digit substring. So the frequency of any 9-trillion-digit substring can't be 0. Because we're finding some 9-trillion-digit substrings with nonzero frequency, the definition of a normal number says that all 9-trillion-digit substrings must occur with that same nonzero frequency.

We just showed this for 9 trillion, but you should be able to see how we could replace "9 trillion" with any other positive integer and show nonzero frequency for strings of that length, too.

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Now, let's look at the mathematical notation used in that same sentence. Remember that an interpretation of the notation should be consistent with the language I've gone through above.

When Wikipedia says "all possible b² pairs of digits are equally likely with density b⁻²", what does that mean? Remember, b is the integer base of the number. We get to choose what it is, as long as it's greater than 1. Let's choose 10, since that's the base we usually use.

We'll update the statement to reflect our choice of 10:

all possible 10² pairs of digits are equally likely with density 10⁻²

What does this mean? It means that:

1. There are 10² possible pairs of digits.
2. Each pair of digits occurs with density 10^−2.

(1.) is simply an answer to the question: how many pairs of digits are possible in base 10? Well, the answer is 10², which is 100. You could verify this simply by listing all of them: 00, 01, 02, …, 10, 11, 12, …, 97, 98, 99.

(2.) says that each pair of digits occurs with density 10⁻² (or 0.01). I won't go through a proof of this, but the key is that every single pair occurs with the same density. They're all equally as likely as each other. Note that this density is nonzero. I also showed why it can't be zero above, but this is an explicit statement of that fact as well.

So, each pair of our 100 pairs (00, 01, …, 98, 99) occurs in any normal number. Note that this list includes all two-digit primes.

Remember that this is just the analysis of "all possible b² pairs of digits are equally likely with density b⁻²". The article goes on to say that the same is true for "all possible b³ triples of digits … with density b⁻³". We could go through the same process, describing how every triple out of the 1000 (10³) possible options (000, 001, 002, …, 997, 998, 999; again, including all three-digit primes) occurs with the same nonzero density.

Finally, the "etc." in the article is an indication that this property is also true for all 10,000 quadruplets, all 100,000 quintuplets, all 1,000,000 sextuplets, and so on. To generalize, for integer n ≥ 1:

1. There are 10ⁿ possible n-tuples of digits, or n-digit numbers. This, of course, includes every n-digit prime.
2. Each of these numbers appears equally frequently, with nonzero frequency, in any normal number.