My posterior probability that you hold two aces should be the same either way
Yes, but the posterior probability is 1⁄3, not 1⁄5.
p(two aces|AH) = 1⁄3 (As the possible options are, AH+AS, AH+2D, AH+2C)
p(two aces|AS) = 1⁄3 (AS+AH, AS+2D, AS+2C)
However, if you had interpreted argument 2 as asking p(two aces|ace of hearts OR ace of spades) you would end up with 1⁄5, which is the same result as the prior p(two aces|have an ace). I think the fallacious reasoning here is that conditioning on the disjunction of having either ace, p(both aces|AH OR AS) = 1⁄5, does not provide you with any new information, as it is the same query as before. But actually selecting one ace and conditioning on that information gives you the correct result, namely that p(both aces|AH) = p(both aces|AS) = 1⁄3. So argument 1 is correct.
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...
My posterior probability that you hold two aces should be the same either way
Yes, but the posterior probability is 1⁄3, not 1⁄5. p(two aces|AH) = 1⁄3 (As the possible options are, AH+AS, AH+2D, AH+2C) p(two aces|AS) = 1⁄3 (AS+AH, AS+2D, AS+2C)
However, if you had interpreted argument 2 as asking p(two aces|ace of hearts OR ace of spades) you would end up with 1⁄5, which is the same result as the prior p(two aces|have an ace). I think the fallacious reasoning here is that conditioning on the disjunction of having either ace, p(both aces|AH OR AS) = 1⁄5, does not provide you with any new information, as it is the same query as before. But actually selecting one ace and conditioning on that information gives you the correct result, namely that p(both aces|AH) = p(both aces|AS) = 1⁄3. So argument 1 is correct.
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...