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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!

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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: $$ \begin{align} &\frac1{52!}\sum_{k=1}^{49}\overbrace{\binom{48}{k-1}}^{\substack{\text{pick $k-1$ from}\\\text{$48$ non-aces}}}\overbrace{\vphantom{\binom11}\ (k-1)!\ }^{\substack{\text{arrange the}\\\text{$k-1$ picked}}}\overbrace{\ \ \ \binom{4}{1}\ \ \ }^{\substack{\text{pick the}\\\text{first ace}}}\overbrace{\ \ \ \binom{3}{1}\ \ \ }^{\substack{\text{pick the}\\\text{second ace}}}\overbrace{\vphantom{\binom11}(51-k)!}^{\substack{\text{arrange the}\\\text{remaining cards}}}\\ &=12\cdot\frac{48!}{52!}\sum_{k=1}^{49}\frac{(51-k)!}{(49-k)!}\\ &=\frac{12}{49\cdot50\cdot51\cdot52}\sum_{k=1}^{49}2\binom{51-k}2\\ &=\frac{12}{49\cdot50\cdot51\cdot52}\sum_{k=1}^{49}2\binom{k+1}2\\ &=\frac{12}{49\cdot50\cdot51\cdot52}\,2\binom{51}3\\[3pt] &=\frac1{13} \end{align} $$

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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.

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  • $\begingroup$ Oh gosh how silly of me, that completely went over my head. Thank you! $\endgroup$ – user107224 Dec 26 '18 at 17:47

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