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wkvong comments on Drawing Two Aces - Less Wrong
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When you say that the posterior probability is 1/3, this depends on the three combinations being equally likely, but as I said in my other comment, they are not equally likely, given your way of obtaining the information.
I see, your solution seems correct now in retrospect. I mistook scenario 2 for being exactly the same as scenario 1, but the two situations where you are not holding the other ace are indeed twice as likely as having both aces (due to selecting the ace at random), so the answer should be 1/5. Looks like I should brush up on my basic probability...