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.
from random import shuffle, randint
flips=1000
HEADS=0
TAILS=1
# individual cells
HMW = HTW = HMD = HTD = 0.0
TMW = TTW = TMD = TTD = 0.0
def run_experiment():
global HMW, HTW, HMD, HTD, TMW, TTW, TMD, TTD
coin = randint(HEADS,TAILS)
if (coin == HEADS):
# wake Beauty on monday
HMW+=1
# drug Beauty on Tuesday
HTD+=1
if (coin == TAILS):
# wake Beauty on monday
TMW+=1
# wake Beauty on Tuesday too
TTW+=1
for i in range(flips):
run_experiment()
print "Total samples where heads divided by total samples ~P(H):",(HMW+HTW+HMD+HTD)/(HMW+HTW+HMD+HTD+TMW+TTW+TMD+TTD)
print "Total samples where woken F(W):",HMW+HTW+TMW+TTW
print "Total samples where woken and heads F(W&H):", HMW+HTW
print "P(W&H)=P(W)P(H|W), so P(H|W)=lim F(W&H)/F(W)"
print "Total samples where woken and heads divided by sample where woken F(H|W):", (HMW+HTW)/(HMW+HTW+TMW+TTW)
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.)
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.
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.