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I mean, yeah two d6 are random but adding them doesn’t give you an even 1-12. It’s a bit more obvious with the d6’s though.

I’m gonna need you to explain your math here. More specifically, I want to know which bytes are going to appear more often when you are generating them in sequence with an equal probability of each digit being zero and one.

Let’s take the other track and instead assume that byte generation is uniformly distributed. In fact, let’s go stronger and assume that any binary number generated by coin flips is uniformly distributed. The base case is a single flip. We have two possibilities, head or tails, each with 50% probability. This means our resulting numbers, 0 and 1, occur with equal probability. This is the uniform distribution. Now assume a binary number of length k - 1 generated by coin flips is uniformly distributed. A binary number of length k is composed of a number of length k - 1 and a single flip. The first k - 1 parts create a range of [0, 2 ^ (k - 1) - 1] and are uniformly distributed. Let’s put our new bit at the front. A 0 gives gives the current range, [0, 2 ^ (k - 1) - 1], and a 1 gives us [2 ^ (k - 1), 2 ^ k - 1]. Note we have an equal probability of falling into either space and they are the same size. In other words, a binary number of length k generated by coin flips is uniformly distributed.