So if I run PSB 20 times, you would assert in each run that P(H|W) = 1⁄21. So now I simply keep you sedated between experiments.
You just changed the problem. If you wake me up between runs of PSB, then P(H|W)=1/21 each time. If not, I have different information to condition on.
I don’t know. It’s a much more complicated problem, because you have 20 coin flips (if I understand the problem correctly). I haven’t taken the time to work through the math yet. It’s not obvious to me, though, why this corresponds to the sleeping beauty problem. In fact, it seems pretty clear that it doesn’t.
The reason it corresponds to Sleeping Beauty is that in the limit of a large number of trials, we can consider blocks of 20 trials where heads was the flip and all values of the die roll occurred, and similar blocks for tails, and have some epsilon proportion left over. (WLLN)
Each of those blocks corresponds to Sleeping Beauty under heads/tails.
You just changed the problem. If you wake me up between runs of PSB, then P(H|W)=1/21 each time. If not, I have different information to condition on.
No; between sedation and amnesia you know nothing but the fact that you’ve been woken up, and that 20 runs of this experiment are to be performed.
Why would an earlier independent trial have any impact on you or your credences, when you can neither remember it nor be influenced by it?
I don’t know. It’s a much more complicated problem, because you have 20 coin flips (if I understand the problem correctly). I haven’t taken the time to work through the math yet. It’s not obvious to me, though, why this corresponds to the sleeping beauty problem. In fact, it seems pretty clear that it doesn’t.
The reason it corresponds to Sleeping Beauty is that in the limit of a large number of trials, we can consider blocks of 20 trials where heads was the flip and all values of the die roll occurred, and similar blocks for tails, and have some epsilon proportion left over. (WLLN)
Each of those blocks corresponds to Sleeping Beauty under heads/tails.