Before she wakes, the probabilities SB would assign if she were conscious are P(monday and heads) = P(monday and tails) = p(tuesday and heads) = p(tuesday and tails) = 1⁄4.
After waking, she would update to p(tuesday and heads) = 0 and P(monday and heads) = P(monday and tails) = p(tuesday and tails) = 1⁄3, since p(tuesday and heads | wakes up) = 0 and p(monday and heads | wakes up) = p(monday and tails | wakes up) = p(tuesday and tails | wakes up) = 1.
Odds(monday and heads : monday and tails : tuesday and heads : tuesday and tails | wakes up)
= Odds(monday and heads : monday and tails : tuesday and heads : tuesday and tails) * Likelihood(wakes up | monday and heads : monday and tails : tuesday and heads : tuesday and tails)
= (1 : 1 : 1 : 1) * (1 : 1 : 0 : 1)
= (1 : 1 : 0 : 1)
= (1/3 : 1/3 : 0 : 1/3)
SB starts out with four equaly likely possibilities. On observing that she wakes up, she eliminates one of them, but does not distinguish between the remaining possiblities. Renormalizing the probabilities gives probability 1⁄3 to the remaining possibilities.
Nope. P(monday and heads)=1/2. P(monday and tails)=1/4. P(tuesday and tails)=1/4. Remember, these have to add to 1.
How come P(monday and heads) and P(monday and tails) are not the same? This is an ordinary unbiased coin, yes?
How come P(monday and tails) and P(tuesday and tails) are not the same. Nothing happens in the interim, yes?
Before she wakes, the probabilities SB would assign if she were conscious are P(monday and heads) = P(monday and tails) = p(tuesday and heads) = p(tuesday and tails) = 1⁄4.
After waking, she would update to p(tuesday and heads) = 0 and P(monday and heads) = P(monday and tails) = p(tuesday and tails) = 1⁄3, since p(tuesday and heads | wakes up) = 0 and p(monday and heads | wakes up) = p(monday and tails | wakes up) = p(tuesday and tails | wakes up) = 1.
Ugh. That makes no sense. Can you explain why she would update in such a manner?
SB starts out with four equaly likely possibilities. On observing that she wakes up, she eliminates one of them, but does not distinguish between the remaining possiblities. Renormalizing the probabilities gives probability 1⁄3 to the remaining possibilities.
I agree, but don’t see how this works as a reply to Phil’s comment.