Probability an ace lies behind first ace Consider a deck of 52 cards. I keep drawing until the first ace appears. I wish to find the probability that the card after is an ace.
Now, the method I know leads to the correct answer is that given the first ace, there are $48$ possible non-ace cards that can be drawn after that. Hence, the probability of drawing another ace is simply $1-\frac{48}{52}=\frac 1{13}$ (credits to @joriki).
However, when trying another method, I got a completely different an bizarre answer. I am inclined to believe that the complement of what is required is that no two aces are consecutive. Hence, we can shuffle the $48$ non-aces ($48!$ total arrangements) with $49$ possible places to insert the aces (which themselves have $4!$ total arrangements). Hence, the probability of the complement should be $\frac{49!}{45!}\frac{48!}{52!}$, and one minus that should give the correct answer. However, this answer is completely off. Where did I go wrong? Thank you!

The original question only asks about the probability of the first 2 aces being consecutive, with no conditions on the other 2 aces (apart from the obvious one that they must be later in the deck).  However, your second method checks for the complement of there not being two consecutive aces anywhere among the 52 cards, which is different, as you found from your result.
