The Lewis Carroll's Pillow Problem's solution was previously addressed here and made me wonder if it makes any sense to reason that in this context the Bayesian framework allows us to estimate "bounds" for the probabilities of the outcomes - in this case the lower bound for the probability of the second counter being white would be 1/2 (if the first counter drawn is the deterministic) and 1 if the first counter drawn was the uncertain one. This way the prior would provide us a way of determining how those limits could be equated to give an estimate.
Assuming that true it seemed reasonable for me to think that this make sense considering that the answer lies in between the bounds (2/3). But this doesn't explain anything about the role of the normalization when we divide the Bayes' numerator by the evidence.
Another aspect of this exercise that I was thinking about was the role of the assumption on our inference process. In this case we assume that we can identify the type of the counter when it is drawn which made think what if that was not the case? (here I am supposing that we could only know which is which after drawing both) Would we still be able to perform Bayesian inference under this scenario or am I mixing up everything incorrectly?
The $\frac23$ can be inferred from the fact that the white can be either the indeterminate, or both the indeterminate and the determinate if the indeterminate was white.
Hence there are $3$ possible draws from the bag, B(W), W(W), (W)W, of which two have the original counter as white, hence the probability is $\frac23$.