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