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RichardKennaway comments on A Proof of Occam's Razor - Less Wrong
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Another way of putting this is that all probability distributions over an infinite enumerated set are biased towards elements that occur earlier, regardless of the principle of enumeration.
And yet another way is this: every computable enumeration of all programs is equivalent to a UTM, and therefore generates the same measure of Kolmogorov complexity, up to the usual additive constant.
Proof: Computability of the enumeration means that there is a program which takes any integer N and returns the Nth in that enumeration. Represent this program as a Turing machine and cascade it with a second Turing machine that executes the program returned by the first TM.
Is this a novel observation? Does it go any distance towards answering Shalizi's question at the end of this note?