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!

An Easier Computation
We can enumerate all of the arrangements where the second ace immediately follows the first with $3$ white markers among $48$ black markers. The first white marker represents the first and second aces, the other white markers represent the other aces; the black markers represent the non-aces. Each of these $\binom{51}{3}$ arrangements of stones represents $48!\,4!$ arrangements of cards where the first ace is followed immediately by the second.
We can enumerate all of the arrangements with $4$ white markers among $48$ black markers. The white markers represent aces; the black markers represent non-aces. Each of these $\binom{52}{4}$ arrangements of stones represents $48!\,4!$ arrangements of cards.
Thus, the probability that the second ace immediately follows the first is
$$
\frac{\binom{51}{3}}{\binom{52}{4}}=\frac{\binom{52}{4}\cdot\frac4{52}}{\binom{52}{4}}=\frac1{13}
$$
For example ('.' = black, ':' = white):
Second ace immediately following the first (the pair represented by first ':'):
......:.......:.........:..........................
represents $6$ non-aces, $2$ aces, $7$ non-aces, $1$ ace, $9$ non-aces, $1$ ace, and $26$ non-aces.
The $4$ aces can be arranged in $4!$ ways and the $48$ non-aces can be arranged $48!$ ways.
Second ace anywhere between the first and third:
......:.......:.........:.......:...................
represents $6$ non-aces, $1$ ace, $7$ non-aces, $1$ ace, $9$ non-aces, $1$ ace, $7$ non-aces, $1$ ace, and $19$ non-aces.
The $4$ aces can be arranged in $4!$ ways and the $48$ non-aces can be arranged $48!$ ways.

Brute Force Computation
I'm a bit slow, so I was not seeing the $1-\frac{48}{52}$ argument clearly, so I applied a sledgehammer.
Summing the probabilities that the first ace is the $k^\text{th}$ card drawn, followed immediately by the second ace:
$$
\begin{align}
&\frac1{52!}\sum_{k=1}^{49}\overbrace{\binom{48}{k-1}}^{\substack{\text{pick $k-1$ from}\\\text{$48$ non-aces}}}\overbrace{\vphantom{\binom11}\ (k-1)!\ }^{\substack{\text{arrange the}\\\text{$k-1$ picked}}}\overbrace{\ \ \ \binom{4}{1}\ \ \ }^{\substack{\text{pick the}\\\text{first ace}}}\overbrace{\ \ \ \binom{3}{1}\ \ \ }^{\substack{\text{pick the}\\\text{second ace}}}\overbrace{\vphantom{\binom11}(51-k)!}^{\substack{\text{arrange the}\\\text{remaining cards}}}\\
&=12\cdot\frac{48!}{52!}\sum_{k=1}^{49}\frac{(51-k)!}{(49-k)!}\\
&=\frac{12}{49\cdot50\cdot51\cdot52}\sum_{k=1}^{49}2\binom{51-k}2\\
&=\frac{12}{49\cdot50\cdot51\cdot52}\sum_{k=1}^{49}2\binom{k+1}2\\
&=\frac{12}{49\cdot50\cdot51\cdot52}\,2\binom{51}3\\[3pt]
&=\frac1{13}
\end{align}
$$
