Morendil comments on Updating, part 1: When can you change your mind? The binary model - Less Wrong
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Right up until your reply prompted me to write a program to check your argument, I wasn't thinking in terms of relative frequencies at all, but in terms of probability distributions.
I haven't learned the rules for relative frequencies yet (by which I mean thing like "(don't) include counts of variables that have a correlation of 1 in your denominator"), so I really have no idea.
Here is my program - which by the way agrees with neq1's comment here, insofar as the "magic trick" which will recover 1/2 as the answer consists of commenting out the TTW line.
However, this seems perfectly nonsensical when transposed to my spreadsheet: zeroing out the TTW cell at all means I end up with a total probability mass less than 1. So, I can't accept at the moment that neq1's suggestion accords with the laws of probability - I'd need to learn what changes to make to my table and why I should make them.
Replying again since I've now looked at the spreadsheet.
Using my intuition (which says the answer is 1/2), I would expect P(Heads, Tuesday, Not woken) + P(Tails, Tuesday, Not woken) > 0, since I know it's possible for Beauty to not be woken on Tuesday. But the 'halfer "variant"' sheet says P(H, T, N) + P(T, T, N) = 0 + 0 = 0, so that sheet's way of getting 1/2 must differ from how my intuition works.
(ETA - Unless I'm misunderstanding the spreadsheet, which is always possible.)
Yeah, that "Halfer variant" was my best attempt at making sense of the 1/2 answer, but it's not very convincing even to me anymore.
That program is simple enough that you can easily compute expectations of your 8 counts analytically.
Your program looks good here, your code looks a lot like mine, and I ran it and got ~1/2 for P(H) and ~1/3 for F(H|W). I'll try and compare to your spreadsheet.