<s>[INST] Drawing without replacement - prob. for an Ace followed by an Ace? Given a standard 52-cards deck:
You are extracting cards from the deck without replacement, until you get an "Ace" for the first time. What is the probability that the next card will be "Ace" too?
I've already seen the following Q&A:
Probability of drawing an Ace: before and after
According to that thread, the answer should be:
 $$\frac{4 \cdot 3}{52 \cdot 51} = \frac{1}{221}$$
But the official solution says that the answer is:
$$\frac{4}{52}$$
which doesn't make sense IMHO. They solved it only with intuition or "mind trick", as they wrote..
My calculation:
Assuming that the 1st card is Ace, then:
$$\frac{4 \cdot 3}{52 \cdot 51} = \frac{1}{221}$$
Assuming that the 2nd card is Ace, then:
$$\frac{(52-4) \cdot 4 \cdot 3}{52 \cdot 51 \cdot 50} = \frac{24}{5525}$$
We notice a pattern here.
Having the 1st Ace at the $k$'th draw, then the probability (for a second Ace after that) is:
$$
p_1 = \frac{ {_{52-4}P_{k-1}} \cdot 4 \cdot 3 }{ {_{52}P_{k-1}} \cdot {_{52-k}P_{2}} }
$$
We need to consider the least-possible scenario - we draw 48 non-Ace cards, then:
$$
p_2 = \frac{48! \cdot 4 \cdot 3}{ {_{52}P_{50}} } = \frac{1}{270725}
$$
So, the required probability is:
$$\begin{align}
p &= p_2 + \sum\limits^{48}_{k=1} p_1 \\
&= p_2 + \sum\limits^{48}_{k=1} \frac{ {_{48}P_{k-1}} \cdot 4 \cdot 3 }{ {_{52}P_{k-1}} \cdot {_{52-k}P_{2}} }\\
&= \frac{1}{270725} + \frac{1696}{20825}\\
&= \frac{1297}{15925}\\
&\cong 0.081444
\end{align}
$$
But my answer is far from either the official solution and from the answer in that thread.
 [/INST] We use a  fairly crude counting approach, in order to rely minimally on intuition. There are $\binom{52}{4}$ equally likely ways to choose the positions of the $4$ Aces. We now count the "favourables." 
Maybe the first two Aces are in positions 1 and 2. That leaves $\binom{50}{2}$ for the rest. 
Maybe they are in positions 2 and 3. That leaves $\binom{49}{2}$ for the rest.
Maybe they are in positions 3 and 4. That leaves $\binom{48}{2}$ for the rest.
And so on, up to positions 49 and 50, leaving $\binom{2}{2}$ for the rest. That gives probability
$$\frac{\binom{50}{2}+\binom{49}{2}+\cdots+\binom{2}{2}}{\binom{52}{4}}.$$
The numerator can be simplified in various ways. My favourite is to count the number of ways to choose $3$ numbers from the first $51$. This can be done in $\binom{51}{3}$ ways.  But the smallest number can be $1$, leaving $\binom{50}{2}$ ways to choose the other two. Or the smallest can be $2$, and so on. We get the number of favourables obtained above.
Our probability is therefore $\frac{\binom{51}{3}}{\binom{52}{4}}$. This simplifies to $\frac{1}{13}$.
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