- #421
Dale
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It isn't. The ratio of the number of days awake is the same as the ratio of the probabilities of being awake.stevendaryl said:Why is the ratio of number of days awake the same as the probability of being awake?
It is given in the problem!stevendaryl said:You said you're assuming P(A|H)/P(A|T) = 1/2. That sounds plausible, but I think it requires proof.
That depends on the sample space you are considering. If the "now" in "what is your credence now" is taken to be a day then yes the frequency is 2 out of 2 days for probability 1. If "now" is one hour then the frequency is 2 out of 48 hours for probability 1/24. If "now" is one minute then the frequency is 2 out of 48*60 ...stevendaryl said:because P(A|T) = 1. (If the coin toss is tails, then she's going to be awake every morning).
In all of those cases, to actually calculate P(A|T) requires assuming information that is not given in the problem. But the ratio P(A|H)/P(A|T) is the same in all those cases and is what is directly given in the problem.