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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!

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  • $\begingroup$ In the first your denominator should be $52 \choose 13$. You have missed multiplying by the number of aces available at each stage, which gives a factor $24$ $\endgroup$ – Ross Millikan May 29 '18 at 16:47
  • $\begingroup$ Thanks for pointing that out! i've changed it now $\endgroup$ – BigWig May 29 '18 at 17:44
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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.

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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

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

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  • 1
    $\begingroup$ And certainly if the multiplication rule weren't required, this application of binomials to "probability of not...." is much cleaner $\endgroup$ – Carl Witthoft May 29 '18 at 19:03
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    $\begingroup$ Or another way to write this (perhaps even simpler): the first ace can go anywhere in the deck, the second ace in 39 places of the remaining 51 possible locations etc. yielding a final result of $1 * 39/51 * 26/50 * 13/49$ $\endgroup$ – soktinpk May 29 '18 at 22:06
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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.

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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

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

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

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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

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

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