edit: I haven’t looked at the post in detail, but I think Morendil did basically the same thing I do here.
Let S count the number of aces of spades you draw, and H the number of aces of hearts. Select one ace randomly from your hand, with equal probability of selection each of the aces you hold. Let R = 1 if the ace of spades is selected, and 0 otherwise. By Baye’s Rule we have:
P(S+H=2 | S+H >= 1 and R=1)= [ P(S+H=2) * P(S+H >= 1 and R=1 | S+H=2) ] / [ P(S+H >= 1 and R=1) ]
P(S+H=2) = 1⁄6 since there are 4 choose 2 = 6 possible hands you can have, and only one has both aces in it.
P(S+H >= 1) = 5⁄6 since there are 6 possible hands, and 5 contain at least 1 ace.
P(R=1|S+H >= 1) = 0 (1/5) + 0 (1/5) + 1 (1/5) + 1 (1/5) + (1/2) * (1/5) = 1⁄2 since there are 5 equiprobable possibilities when S+H >= 1. Two of them occur when you don’t have the ace of spades, so P(R=1)=0, while two occur when do have the ace of spades but not the ace of hearts, so P(R=1)=1. Finally, when you have both, P(R=1)=1/2.
P(S+H >= 1 and R=1 | S+H=2) = P(R=1 | S+H = 2) = 1⁄2 by the definition of R.
In some detail:
edit: I haven’t looked at the post in detail, but I think Morendil did basically the same thing I do here.
Let S count the number of aces of spades you draw, and H the number of aces of hearts. Select one ace randomly from your hand, with equal probability of selection each of the aces you hold. Let R = 1 if the ace of spades is selected, and 0 otherwise. By Baye’s Rule we have:
P(S+H=2) = 1⁄6 since there are 4 choose 2 = 6 possible hands you can have, and only one has both aces in it.
P(S+H >= 1 and R=1) = P(S+H >= 1)*P(R=1|S+H >= 1) and:
P(S+H >= 1 and R=1 | S+H=2) = P(R=1 | S+H = 2) = 1⁄2 by the definition of R.
Putting it all together we have:
Which is to say, I’ll go with argument 2.
Alternatively, argument 1 is wrong because the state of knowledge is not the same as in scenario 1.