Shahrazad is put to sleep on Sunday night. Depending on the hidden toss of a fair coin, she is awaken either once (Heads) or twice (Tails). After each awakening, she gets back to sleep and forget that awakening. When awakened, what should her probability of Heads be?
My first reaction is to argue that Shahrazad does not gain information between the time she goes to sleep when the coin is fair and the time(s) she is awaken, apart from being awaken, since she does not know how many times she has been awaken, so the probability of Heads remains ½. However, when thinking more about it on my bike ride to work, I thought of the problem as a decision theory or betting problem, which makes ⅓ the optimal answer.
I then read [if not the huge literature] a rather extensive analysis of the paradox by Ciweski, Kadane, Schervish, Seidenfeld, and Stern (CKS³), which concludes at roughly the same thing, namely that, when Monday is completely exchangeable with Tuesday, meaning that no event can bring any indication to Shahrazad of which day it is, the posterior probability of Heads does not change (Corollary 1) but that a fair betting strategy is p=1/3, with the somewhat confusing remark by CKS³ that this may differ from her credence. But then what is the point of the experiment? Or what is the meaning of credence? If Shahrazad is asked for an answer, there must be a utility or a penalty involved otherwise she could as well reply with a probability of p=-3.14 or p=10.56… This makes for another ill-defined aspect of the “paradox”.
Another remark about this ill-posed nature of the experiment is that, when imagining running an ABC experiment, I could only come with one where the fair coin is thrown (Heads or Tails) and a day (Monday or Tuesday) is chosen at random. Then every proposal (Heads or Tails) is accepted as an awakening, hence the posterior on Heads is the uniform prior. The same would not occurs if we consider the pair of awakenings under Tails as two occurrences of (p,E), but this does not sound (as) correct since Shahrazad only knows of one E: to paraphrase Jeffreys, this is an unobservable result that may have not occurred. (Or in other words, Bayesian learning is not possible on Groundhog Day!)