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...
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...