So, I'm still working on this in my plodding, newbie-at-probability-math fashion.

What I took away from my exchanges with AlephNeil is that I get the clearest picture if I think in terms of a joint probability distribution, and attempt to justify mathematically each step of my building the table, as well as the operations of conditioning and marginalizing.

In the original Sleeping Beauty problem, we have three variables: x is how the coin came up {heads, tails}, y is the day of the week {monday, tuesday}, and z is whether I am asked for my credence (i.e. woken) {wake, sleep}.

P(x,y,z)=P(x)P(y|x)P(z|x,y) and unlike in the "revival" case x and y aren't clearly independent. In fact the answer very much seems to hinge on what we take the probability of it being tuesday, given that the coin came up heads.

The relevant possible outcomes are: (H,M,W) (H,T,W) (T,M,W) (T,T,W) (H,M,S) (H,T,S) (T,M,S) (T,T,S) - eight in all.

Conditioning on z=W consists of deleting the part of the table that has z=S, summing up all the remaining values, and renormalizing by dividing every cell in cell in the table by the total.

The rules for filling the table are: the values must add up to 1; the "heads" and "tails" branches must receive equal probability mass from P(x); and P(z|x,y) must reflect the experimental rules. So we must have the following:

• P(H,M,W) - see below
• P(H,T,W)=0
• P(T,M,W)=1/4
• P(T,T,W)=1/4
• P(H,M,S)=0
• P(H,T,S) - see below
• P(T,M,S)=0
• P(T,T,S)=0

The ambiguity seems to arise in allocating probability mass to the outcomes: "the coin comes up heads; it is Monday; I get woken up", and "the coin comes up heads; it is Tuesday; I do no get woken up". That is, I'm not sure what the correct conditional distribution P(y|x) should be.

The 1/2 answer corresponds to allocating all of the available 1/2 probability mass to the first of these outcomes in the joint table, saying P(y=M|x=H)=1 and P(y=T|x=H)=0. Or verbally, "it's certain that I get woken up on Monday if the coin comes up heads, and after that the experiment is over". The "not woken up" half of the table receives no probability mass at all.

The 1/3 answer corresponds to distributing that probability mass among the two outcomes, saying P(y=M|x)=P(y=T|x)=1/2. Verbally: "however the coin comes up, it could be either Monday or Tuesday". Here 1/4 of the total probability mass is in the "not woken up" half of the table and gets deleted when we condition on being woken.

(ETA: Where does the amnesia appear in this formalization? It doesn't, but neither does it need to. Its only practical consequence is to outlaw conditioning on the day, so working out the distribution P(x|z) conforms to the amnesia.)