All of Flumber's Comments + Replies

Flumber10

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.)

-1dellbarnes
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.)