(EDITED) Computing this the long way, by straightforward application of the product rule, yields probability 1/5.

There is a shortcut corresponding to argument 2. Call H the hypothesis that you have two aces, and E the evidence constituted by my confirming the ace I pick at random is spades. It turns out that P(E|H)=P(E|!H), so the evidence can't change my probability assignment. P(E|H) = P(spades | 2 aces) = 1/2. P(E|!H) = P (spades | 1 ace) = 1/2.

Now do it the long way: I start from P(2A,SR|A), that is, the probability that I hold two aces AND that I chose one at random, getting the ace of spaces, GIVEN that I hold one ace.

By the product rule P(2A,SR|A)=P(2A|SR,A)P(SR|A)=P(SR|2A,A)P(2A|A) and therefore the answer we seek is P(SR|2A,A)P(2A|A)/P(SR|A).

The numerator is easy. P(SR|2A,A)=P(SR|2A)=1/2 and P(2A|A) is as in scenario 2.

The denominator is trickier, requiring one more application of the product rule to get P(SR|A)=P(SR,A)/P(A) and factoring A into the six mutually exclusive possibilities to get P(SR,A). Of these the three which matter are (AS,2C),(AS,2D),(AS,AH) - for these P(SR)=1 for the first two and 1/2 for the last.

Numerically, P(SR|A) comes out to 1/2, which cancels out with the other 1/2, leaving the answer of 1/5. The reason they cancel out is precisely the "shortcut" above: P(SR|A)=P(SR|2A).