I wrote up a response to this, but I thought it was also worthwhile writing a comment that directly responds to the argument about whether we can update on a random bit of information.
@travisrm89 wrote:
How can receiving a random bit cause Beauty to update her probability, as in the case where Beauty is an AI? If Beauty already knows that she will update her probability no matter what bit she receives, then shouldn’t she already update her probability before receiving the bit?
Ksvanhorn responds by pointing out that this assumes that the probabilities add to one, while we are considering the probability of observing a particular sequence at least once, so these probabilities overlap.
This doesn’t really clarify what is going on, but I think that we can clarify this by first looking at the following classical probability problem:
A man has two sons. What is the chance that both of them are born on the same day if at least one of them is born on a Tuesday?
Most people expect the answer to be 1⁄7, but the usual answer is that 13⁄49 possibilities have at least one born on a Tuesday and 1⁄49 has both born on Tuesday, so the chance in 1⁄13. Notice that if we had been told, for example, that one of them was born on a Wednesday we would have updated to 1⁄13 as well. So our odds can always update in the same way on a random piece of information if the possibilities referred to aren’t exclusive as Ksvanhorn claims.
However, consider the following similar problem:
A man has two sons. We ask one of them at random which day they were born and they tell us Tuesday. What is the chance that they are both born on the same day?
Here the answer is 1⁄7 as we’ve been given no information about when the other child was born. When Sleeping Beauty wakes up and observes a sequence, they are learning that this sequence occurs on a on a random day out of those days when they are awake. This probability is 1/n where n is the number of possibilities. This is distinct from learning that the sequence occurs in at least one wakeup just like learning a random child is born on a Tuesday is different from learning that at least one child was born on a Tuesday. So Ksvanhorn has calculated the wrong thing.
When Sleeping Beauty wakes up and observes a sequence, they are learning that this sequence occurs on a on a random day out of those days when they are awake.
That would be a valid description if she were awakened only on one day, with that day chosen through some unpredictable process. That is not the case here, though.
What you’re doing here is sneaking in an indexical—“today” is either Monday if Heads, and “today” is either Monday or Tuesday if Tails. See Part 2 for a discussion of this issue. To the extent that indexicals are ambiguous, they cannot be used in classical propositions. The only way to show that they are unambiguous is to show that there is an equivalent way of expressing that same thing that doesn’t use any indexical, and only uses well-defined entities—in which case you might as well use the equivalent expression that has no indexical.
I wrote up a response to this, but I thought it was also worthwhile writing a comment that directly responds to the argument about whether we can update on a random bit of information.
@travisrm89 wrote:
Ksvanhorn responds by pointing out that this assumes that the probabilities add to one, while we are considering the probability of observing a particular sequence at least once, so these probabilities overlap.
This doesn’t really clarify what is going on, but I think that we can clarify this by first looking at the following classical probability problem:
Most people expect the answer to be 1⁄7, but the usual answer is that 13⁄49 possibilities have at least one born on a Tuesday and 1⁄49 has both born on Tuesday, so the chance in 1⁄13. Notice that if we had been told, for example, that one of them was born on a Wednesday we would have updated to 1⁄13 as well. So our odds can always update in the same way on a random piece of information if the possibilities referred to aren’t exclusive as Ksvanhorn claims.
However, consider the following similar problem:
Here the answer is 1⁄7 as we’ve been given no information about when the other child was born. When Sleeping Beauty wakes up and observes a sequence, they are learning that this sequence occurs on a on a random day out of those days when they are awake. This probability is 1/n where n is the number of possibilities. This is distinct from learning that the sequence occurs in at least one wakeup just like learning a random child is born on a Tuesday is different from learning that at least one child was born on a Tuesday. So Ksvanhorn has calculated the wrong thing.
That would be a valid description if she were awakened only on one day, with that day chosen through some unpredictable process. That is not the case here, though.
What you’re doing here is sneaking in an indexical—“today” is either Monday if Heads, and “today” is either Monday or Tuesday if Tails. See Part 2 for a discussion of this issue. To the extent that indexicals are ambiguous, they cannot be used in classical propositions. The only way to show that they are unambiguous is to show that there is an equivalent way of expressing that same thing that doesn’t use any indexical, and only uses well-defined entities—in which case you might as well use the equivalent expression that has no indexical.