An ordinary deck of $52$ playing cards is randomly divided into $4$ piles of $13$ cards each. Compute the probability that each pile has exactly one ace.
The answer provided is is $(39*26*13)/(51*50*49) \approx 0.105$
The above answer uses conditional probability, but I would like to know what's wrong with my reasoning:
Call the four piles of partitions $1$, $2$, $3$, and $4$.
- For partition $1$, there are ${4 \choose 1}$ ways to choose which ace the partition will contain. Then, there are ${48\choose 12}$ ways to choose the remaining $12$ cards, as we cannot choose any other aces.
- For partition $2$, there are ${3\choose 1}$ ways to choose which ace the partition will contain. Then, there are ${36\choose 12}$ ways to choose the remaining $12$ cards, as there are only $36$ non-ace cards left.
Following similar reasoning for partitions $3$ and $4$, we find that there are ${2\choose 1}{24 \choose 12}$ and ${1\choose 1}{12 \choose 12} = 1$ ways to form those hands.
Therefore, my probability is given by
$$\frac{4 \cdot {48\choose 12} + 3\cdot{36\choose 12} + 2\cdot{24\choose 12}}{{52\choose 13}{39\choose 13}{26\choose 13}} \not \approx 0.105$$
The denominator is the number of ways to choose the cards in each hand without any constraints.
I am not sure what is wrong with my computation.
Best Answer
You were close. You just have to multiply the possibilities in the numerator: $$\frac{4 \cdot {48\choose 12} \cdot 3\cdot{36\choose 12} \cdot 2\cdot{24\choose 12}}{{52\choose 13}{39\choose 13}{26\choose 13}} \approx 0.105$$