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

Question

I'm trying to answer this question and you are supposed to use the multiplication rule to solve it:

A deck of 52 playing cards is randomly divided into four piles of 13 cards each. Compute the probability that each pile has exactly 1 ace.

I started off by defining following 4 events: $A_{1}, A_{2}, A_{3}$ and $A_{4}$ where $A_{i}$ denotes the event that exactly one ace is found in the i$^{th}$ pile - so to find the probability I need to find the probability of the intersection of all these events which is where I can use the multiplication rule.

The multiplication rule says that $$\mathbb{P}(A_{1} \cap A_{2} \cap A_{3} \cap A_{4})= \mathbb{P}(A_{1}) \mathbb{P}(A_{2}|A_{1}) \mathbb{P}(A_{3}|A_{2} \cap A_{1}) \mathbb{P}(A_{4}|A_{3} \cap A_{2} \cap A_{1})$$

to find each of the probabilities on the RHS I looked compared the possible combinations allowed for each situation:

The number of possible card combinations such that $A_{1}$ holds is $\binom{48}{12}$ and the total number of possible card combinations for the first pile is $\binom{52}{13}$. It then follows that $$\mathbb{P}(A_{1})= \frac{\binom{48}{12}}{\binom{52}{13}}=\frac{1406}{4165}$$

When moving on to the second pile, it follows that we now have 39 cards remaining so in order for the pile 2 to have exactly one ace, it leads to $\binom{36}{12}$ possible combinations out of the $\binom{39}{13}$ total number of combinations and so we get that $$\mathbb{P}(A_{2}|A_{1}) = \frac{\binom{36}{12}} {\binom{39}{13}} =\frac{225}{703}$$.

Continuing on in this way I got that

$$\mathbb{P}(A_{3}|A_{2} \cap A_{1}) = \frac{\binom{24}{12}}{\binom{26}{13}}=\frac{13}{50}$$

and by the way I have defined my events, it means that $$\mathbb{P}(A_{4}|A_{3} \cap A_{2} \cap A_{1}) = 1$$

so by the multiplication rule I get that $$\mathbb{P}(A_{1} \cap A_{2} \cap A_{3} \cap A_{4}) = \frac{1406}{4165} \frac{225}{703} \frac{13}{50} \approx 0.0281$$

However the answer I am given says it should be $\approx 0.105$. Can anyone help me to see where I have gone wrong? Would it perhaps be that defining the events differently lead to different probabilities? Thanks!

Answer 1

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.

Answer 2

You have ignored the fact that there are $4$ different aces, so $P(A_1) = \dfrac{4\binom{48}{12}}{\binom{52}{13}}$, etc.

Answer 3

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.

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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

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

Answer 4

A different approach:

There are

-- $\binom{4}{1}$ ways of selecting an ace and $\binom{48}{12}$ ways of selecting other cards to the $1$st deck, while $\binom{52}{13}$ doing so in total;

-- $\binom{3}{1}$ ways of selecting an ace and $\binom{36}{12}$ ways of selecting other cards to the seond deck, while $\binom{39}{13}$ doing so in total;

-- $\binom{2}{1}$ ways of selecting an ace and $\binom{24}{12}$ ways of selecting other cards to the third deck, while $\binom{36}{13}$ doing so in total;

-- this one is easy.

Thus, we have

$$\frac{\binom{4}{1}\binom{48}{12}}{\binom{52}{13}}\times\frac{\binom{3}{1}\binom{36}{12}}{\binom{39}{13}}\times \frac{\binom{2}{1}\binom{24}{12}}{\binom{26}{13}} \approx 0.1055.$$

Answer 5

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.

Answer 6

Choose $12$ non-ace cards for the first pile, $12$ non-ace cards for the second pile, $12$ non-ace cards for the third pile and the remaining $12$ non-ace cards for the fourth pile.

Place the four aces in the four piles, one in each pile.

This can be done in $${48 \choose 12}{36 \choose 12}{24 \choose 12}{12 \choose 12}\cdot 4!$$ ways.

This has to be divided with the number of all possibilities: choose $13$ cards for the first pile, $13$ for the second, $13$ for the third and the remaining $13$ for the fourth. This can be done in

$${52 \choose 13}{39 \choose 13}{26 \choose 13}{13 \choose 13}$$

ways.

The result is

$$\frac{{48 \choose 12}{36 \choose 12}{24 \choose 12}{12 \choose 12}\cdot 4!}{{52 \choose 13}{39 \choose 13}{26 \choose 13}{13 \choose 13}} = 0.1055$$