Argument 2 is correct, but I'm having considerably difficulty putting the reason into succinct terms in a way that'd feel satisfying for me, even after reading all the comments here.

However, here's the thoroughly worked out answer for anyone who wants it:

We start with no questions asked. All of the six possible sets are equiprobable.

• (AS 2C): 1/6
• (AS 2D): 1/6
• (AS AH): 1/6
• (AH 2C): 1/6
• (AH 2D): 1/6
• (2C 2D): 1/6

Next you ask the first question, "do you have an ace?" I respond "yes". This eliminates the (2C 2D) possibility. This leaves the following sets:

• (AS 2C) or (AS 2D): 2/5
• (AH 2C) or (AH 2D): 2/5
• (AS AH): 1/5

Now the probability of me holding both aces is 1/5, the probability that I have (only) the Ace of Spades is 2/5 (3/5), and the probability that I have (only) the Ace of Hearts is 2/5 (3/5).

Suppose that after this, you're about to ask me "is a randomly picked ace the ace of spades?" The prior probability that I'll answer "yes" is, for the different scenarios:

• (AS 2C) or (AS 2D): 2/5 * 1 = 2/5
• (AH 2C) or (AH 2D): 2/5 * 0 = 0
• (AS AH): 1/5 * 1/2 = 1/10

Totaling 5/10, or 1/2.

Assuming that I have answered "yes", that leaves three possible sets:

• (AS 2C)
• (AS 2D)
• (AS AH)

However, they are not equiprobable, as there was previously a random element involved. Now that we know I have the Ace of Spades, let's take another look at the probabilities for "is a randomly picked ace the ace of spades?"

• (AS 2C): 1/3 * 1 = 1/3
• (AS 2D): 1/3 * 1 = 1/3
• (AS AH): 1/3 * 1/2 = 1/6

Totaling 5/6. So now, suppose you only know that I have an ace. What is the probability that I have both aces, given that a randomly chosen ace from my hand produced the Ace of Spades?

By Bayes' rule, P(A|B) = P(B|A)P(A) / P(B). In this case, A = I have both aces, B = A randomly chosen ace from my hand produces the Ace of Spades. Naturally, that makes B|A = A randomly chosen ace from my hand produces the Ace of Spades, given that I have both aces.

P(A) = P(I have both aces) = 1/5

P(B) = P(A randomly chosen ace from my hand produces the Ace of Spades) = 1/2

P(B|A) = P(A randomly chosen ace from my hand produces the Ace of Spades, given that I have both aces) = 1/2

Thus P(A|B) = P(B|A)P(A) / P(B) = (1/2)(1/5) / (1/2) = (1/10) / (1/2) = 1/5.