# Finding the probability that an ace is found in every pile when a deck of cards is split into 4

ETA: I see that the use of the multiplication rule is expected. I'm still going to leave this here as an approach to use for this kind of problem, in general.

Alternate approach. Place the other $48$ cards into four piles, accounting for order; there are $48!$ ways to do this. Then assign each of the four aces to one of the piles; there are $4!$ ways to do this. Finally, each ace has to be placed into one of $13$ different positions in its respective pile; there are $13^4$ ways to do this.

Since there are $52!$ different arrangements of the deck overall, the desired probability is

$$\frac{13^4 \times 4! \times 48!}{52!} = \frac{2197}{20825} \approx 0.10550$$

First, a small typo:

$$\mathbb{P}(A_{1})= \frac{\binom{48}{12}}{\binom{48}{12}}=\frac{1406}{4165}$$

I assume you meant:

$$\mathbb{P}(A_{1})= \frac{\binom{48}{12}}{\binom{52}{13}}$$

However, when I calculate that, I get

$$\frac{\binom{48}{12}}{\binom{52}{13}} = \frac{9139}{83300}$$

More importantly, since there are $4$ aces to choose from for pile $1$, it really should be:

$$\mathbb{P}(A_{1})= \frac{4 \cdot \binom{48}{12}}{\binom{52}{13}}=\frac{9139}{20825}$$

Likewise, something went wrong with your calculation here:

$$\mathbb{P}(A_{2}|A_{1}) = \frac{\binom{36}{12}} {\binom{39}{13}} =\frac{225}{703}$$

When I calculate that, I get

$$\frac{\binom{36}{12}} {\binom{39}{13}} = \frac{325}{2109}$$

But again, more importantly, since there are $3$ aces left to choose from, it really should be:

$$\mathbb{P}(A_{2}|A_{1})=\frac{3\cdot \binom{36}{12}} {\binom{39}{13}} =\frac{325}{703}$$

And likewise, since there are $2$ aces left for pile $3$, it should be:

$$\mathbb{P}(A_{3}|A_{2} \cap A_{1}) = \frac{2 \cdot \binom{24}{12}}{\binom{26}{13}}=\frac{13}{25}$$

And so, we get:

$$\mathbb{P}(A_{1} \cap A_{2} \cap A_{3} \cap A_{4}) = \frac{9139}{20825} \frac{325}{703} \frac{13}{25} \approx 0.1055$$

As desired. So, you did the basic method largely correct, but you did some sloppy calculations, and more importantly, you forgot to take into account that for the first few piles you have a choice of aces.

Another approach using the multiplication rule

The multiplication rule can be very simply applied imagining $52$ slots divided into $4$ piles of $13$

$\small{\square\square\square\square\square\square\square\square\square\square\square\square\square\quad\square\square\square\square\square\square\square\square\square\square\square\square\square\quad\square\square\square\square\square\square\square\square\square\square\square\square\square\quad\square\square\square\square\square\square\square\square\square\square\square\square\square}$

The $1^{st}$ ace can go to any slot, and given that, the $2^{nd}$ ace has $39$ out of $51$ free slots to go in,
and so on and so forth, thus simply$\quad\frac{39}{51}\cdot\frac{26}{50}\cdot\frac{13}{49}$

This way, we need not bother at all as to how the other $48$ cards are distributed.