If we want to replicate the situation 1000 times, we shouldn't end up with 1500 observations. The correct way to replicate the awakening decision is to use the probability tree I included above. You'd end up with expected cell counts of 500, 250, 250, instead of 500, 500, 500.

Beauty ends up with 1500 observations on average (maybe as few as 1000 or as many as 2000). Imagine a sequence of Beauty-observations in (H|TT)^1000 , where by r^1000 I mean 1000 repetitions of r. This string is from 1000-2000 letters long.

If you consider the scenario from a non-amnesiac perspective, then you can consider the TT - the two forgetful-Beauty observations in the tails case, as a single event, which is indeed equally likely to the alternative, H. In fact, the shortest possible coding to describe one of the beauty-observation strings is just a 1000-bit string where the nth bit indicates the result of the nth coin flip.

But what are you thinking when you say there are two "cells" each with p=1/4 (count 250 out of 1000)? What, exactly, would happen to 250 times on average? Certainly we expect Beauty waking on Mon. with it being tails 500 times (and also 500 times on Tue.).