The first reply eliminates the no-ace case from six equally likely cases, leaving two aces as one of five equally likely cases. So the probability is one fifth. (The second question is irrelevant, by symmetry.)
If we have an ace in the hand, 2 of those “equally likely cases” are no longer possible. (2 of those cases involve the other ace and a non-ace card.)
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The first reply eliminates the no-ace case from six equally likely cases, leaving two aces as one of five equally likely cases. So the probability is one fifth. (The second question is irrelevant, by symmetry.)
If we have an ace in the hand, 2 of those “equally likely cases” are no longer possible. (2 of those cases involve the other ace and a non-ace card.)