It seems that most readers swallowed my arguments in the previous post, so I think it is time to turn the screw a few more times so to speak.
Again, we are confronted with a deck of 2 cards (we know that there are only black and red cards, no jokers), we take the top card and it is black. What is the probability that the remaining card is also black?
In the previous post we only considered two hypotheses (*):
H1 ... black cards only
H2 ... mixed deck
and we assumed p(b|H2) = 1/2.
This assumption bothers me now.
We don't know how the 2-card deck was put together and it could be that somebody made sure we would always find a black card on top, even for the mixed deck. So I think we need to split H2 into two different hypotheses.
H2a ... manipulated mixed deck: p(b|H2a) = 1
H2b ... random mixed deck: p(b|H2b) = 1/2
Again, relying on the principle of indifference we use an uninformed prior p(H1) = p(H2a) = p(H2b) = 1/3 and
find p(H1|b) = 2/5 (I recommend that you actually plug in the numbers and do the calculation).
In other words, the probability that the remaining card is also black (2/5) is now significantly less than
the probability that it is red (3/5).
But it is clear that we have not yet achieved the goal of "absolutely no information" in selecting our prior; The problem is H2b.
We have to split it further to take into consideration cases in between outright
manipulation and random selection. So we have to consider additional N hypotheses
H2k ... k=1 ... N with p(b|H2k) = k/N
and let N go to infinity (o).
As you can easily check, the large number of hypotheses for mixed decks results in p(H1|b) -> 0 and therefore
we have to conclude that using an absolutely uninformed prior the probability that the
remaining card is black is (close to) zero.
This remarkable result is due to the fact that there is only one way how the deck can consist of only black cards,
but there are many ways how a mixed deck could have been put together. Absolutely no information means certainty about the 2nd card in this case, thanks to the power of the Bayesian method (x).
(*) Nobody complained, but it was a bit sloppy to exclude H0 = 'red only' from the prior (which should be chosen before the black card was seen). But p(b|H0) = 0 and, as you can check, including H0 would have made no difference.
(o) The limit N = infinite would require the use of an improper prior and I think it is sufficient for our purpose to consider the case of large but finite N.
(x) I think this toy model will be very helpful e.g. in cosmology and multiverse research.