The conclusion of the second is correct, but to arrive at that conclusion I had to write out all the possibilities and observe how the sequence of answers pruned them. I only understood the second argument when I realised that the symmetry of spades and hearts is what makes it work.
By that symmetry, the posterior probability of having two aces after answering yes to the final question about the ace of spades—P(2A|AS=yes) for short—must equal P(2A|AH=yes). But P(2A|AH=yes) = P(2A|AS=no). So the posterior after asking about the ace of spades is independent of the answer, therefore etc.
If we add the ace of wands to the pack, I make the probability of having two aces in each scenario to be S1: 1⁄2, S2: 2⁄9, S3: 1⁄3. At the end of S3, there are the same four possibilities remaining as in S1, but they are not equiprobable: AS+AH and AS+AW have had half their probability mass pruned away by the random selection of an ace.
The conclusion of the second is correct, but to arrive at that conclusion I had to write out all the possibilities and observe how the sequence of answers pruned them. I only understood the second argument when I realised that the symmetry of spades and hearts is what makes it work.
By that symmetry, the posterior probability of having two aces after answering yes to the final question about the ace of spades—P(2A|AS=yes) for short—must equal P(2A|AH=yes). But P(2A|AH=yes) = P(2A|AS=no). So the posterior after asking about the ace of spades is independent of the answer, therefore etc.
If we add the ace of wands to the pack, I make the probability of having two aces in each scenario to be S1: 1⁄2, S2: 2⁄9, S3: 1⁄3. At the end of S3, there are the same four possibilities remaining as in S1, but they are not equiprobable: AS+AH and AS+AW have had half their probability mass pruned away by the random selection of an ace.