If you play bridge, and are serious, you learn a principle called ‘restricted choice’ that works the same way. Suppose that I am missing both the king and queen in a key suit. I do not know which opponent (LHO or RHO) has it. Suppose I play the jack, and RHO wins it with the king. Now LHO is a 2 – 1 favorite to hold the queen. The reason is that if RHO had the king only, he would always play the king — likelihood of 1. If he had both the K and Q, he would play the K only half the time — likelihood of 1/2. This skews the posteriors toward LHO holding the Q.

As some of the others have pointed out, the problem here is that you don’t just learn that Alice has a red ace, but also that she said she has a red ace. Similarly, in the Monty Hall problem, you don’t just learn that door 3 has no prize, but that Monty has shown that door 3 has no prize. In each case, one of the situations (the case where Alice has both aces, the case where the door you selected in the first round has the prize) allows two different ways for Alice/Monty to reveal the information. Thus, on some specifications of what happens, each piece of information only has a 50% chance in this situation, and thus in some sense, half of this possibility is excluded as well when the information is received. When you work through the calculations, you see that the proper credence both before and after this extra piece of information is still 1/5 in the Alice case and 1/3 in the Monty case.

It looks like the puzzle really arises from two incompatible ways of treating the state space. The official way it was given in the statement of the puzzle treated the two positions in Alice’s hand as indistinguishable. But in that case there’s no way to talk of the color of “that ace”; i.e, there’s no proposition in the state space to be expressed by “That ace is black”, as opposed to just “I’m holding a black ace.” But if Alice only tells Bob “I’m holding a black ace”, then Bob’s second inference is mistaken. As Joel Pust explained, and are both still 1/5.

The alternative is that the two positions in Alice’s hand are distinguished—say she’s holding one of them in her left hand and the other in her right hand. In that case there are in fact 12 different atomic states: , and so on. Then if Alice announces “The card in my left hand is an ace”, there are six possibilities remaining, and the second card is an ace in two of those six possibilities. So the probability after Alice’s first announcement that she is holding a second ace is 1/3. Her later announcement of its color won’t change this.

So there are two pieces of reasoning in the puzzle, one that leads to 1/5, another that leads to 1/3. Only the first is right if the cards are indistinguishable; only the second is right if the cards are distinguishable.

My reasoning was as follows. Suppose the card is announced red, then Bob has just found out that Alice has (at least) the ace of hearts, i.e. Bob finds out that AH. Now

But now I see that the way that I presented the problem doesn’t vindicate this. Indeed, as the problem is presented here, if the card is announced red, Bob learns from Alice, not just the proposition that she has the ace of hearts, but also that she asserts this. Let stand in for ‘Alice says that P‘. We get, for example:

I think that we probably get the same kind of result even if we do away with Alice’s assertion altogether, so that Bob finds things out ‘directly’, so to speak.