I didn’t read either argument before cranking up Bayes’ Theorem. After the first question, the five remaining possibilities are equiprobable, so the posterior probabilities are proportional to the likelihoods of each possibility. The two non-pairs containing the ace of spades have likelihood 1 and the pair of aces has likelihood 0.5. Normalizing the likelihoods gives 1⁄5 chances for the pair of aces.
I didn’t read either argument before cranking up Bayes’ Theorem. After the first question, the five remaining possibilities are equiprobable, so the posterior probabilities are proportional to the likelihoods of each possibility. The two non-pairs containing the ace of spades have likelihood 1 and the pair of aces has likelihood 0.5. Normalizing the likelihoods gives 1⁄5 chances for the pair of aces.