Not all of the waking moments have the same probability of occurring. If you estimate the probability of heads by the proportion of waking moments that were preceded by heads, you’d be throwing out information. Again, on a random waking moment, Monday preceded by heads is more likely than Monday preceded by tails.
On a random waking moment, Monday preceded by heads is equally likely as Monday preceded by tails.
I think you’re thinking of a similar problem that we discussed last year, which involves a forgetful driver who is driving past 1 to n intersections, and needs to turn left at at least one of them. That problem is different, because it’s asking about the probability of turning left at least once over the course of his drive.
The absent-minded driveris essentially the same problem, but it’s easier to analyze because explicit payoff specification prompts you to estimate expected value of possible strategies. In estimating those strategies, we use the same probability model that would say “1/2” in the Beauty problem.
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.
Not all of the waking moments have the same probability of occurring. If you estimate the probability of heads by the proportion of waking moments that were preceded by heads, you’d be throwing out information. Again, on a random waking moment, Monday preceded by heads is more likely than Monday preceded by tails.
On a random waking moment, Monday preceded by heads is equally likely as Monday preceded by tails.
I think you’re thinking of a similar problem that we discussed last year, which involves a forgetful driver who is driving past 1 to n intersections, and needs to turn left at at least one of them. That problem is different, because it’s asking about the probability of turning left at least once over the course of his drive.
The absent-minded driver is essentially the same problem, but it’s easier to analyze because explicit payoff specification prompts you to estimate expected value of possible strategies. In estimating those strategies, we use the same probability model that would say “1/2” in the Beauty problem.
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.