# 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.