A probability problem by Ujjwal Mani Tripathi

There are $\frac{52!}{13! \times 13! \times 13! \times 13!}$ ways of splitting the pack into $4$ piles of $13$ , and there are $4! \times \frac{48!}{12! \times 12! \times 12! \times 12!}$ ways of splitting the pack so that there is one Ace in each pile (there are $4!$ ways of distributing the Aces amongst the piles, and then $\frac{48!}{(12!)^4}$ ways of distributing the remaining cards amongst the four piles).

Thus the probability that there is one Ace in each pile is $\frac{4! \times 13^4}{49 \times 50 \times 51 \times 52} \; = \; \frac{2197}{20825}$ making the answer $\boxed{23022}$ .

Define events $\text{ E\_i such that i} = 1 , 2 , 3 , 4 \text{ as follows }$ .

$\displaystyle E_1 =$ the ace of spades is in any one of the piles .

$E_2 =$ the ace of spades and the ace of hearts are in different piles

$E_3 =$ the ace of spades , hearts and diamonds are all in different piles

$E_4 =$ all $4$ aces are in different piles

The probability desired is $P(E_1E_2E_3E_4)$ and by multiplication rule

$P(E_1E_2E_3E_4) = P(E_1) * P(E_2 | E_1) * P(E_3 | E_2E_1) * P(E_4 | E_3E_2E_1)$

Now , $P(E_1) = 1$ , since $E_1$ is the sample space $S$ .

$P(E_2 | E_1) = \dfrac{39}{51}$ , since the pile containing the ace of spades will receive $12$ out of remaining $51$ cards .

$P(E_3 | E_2E_1) = \dfrac{26}{50}$ , since the pile containing the ace of spades and hearts will receive $24$ of the remaining $50$ cards ; and finally...

$P(E_4 | E_3E_2E_1) = \dfrac{13}{49}$

Therefore , we obtain that the probability that each pile has exactly $1$ ace is

$P(E_1E_2E_3E_4) = \dfrac{39*26*13}{51*50*49} = \dfrac{2197}{20825}$

Thus the final answer becomes $\large = 23022$

Let's consider the aces as identical cards. If we display the 52 cards along a line, there are ${52 \choose 4}$ ways to place the 4 identical aces.

Now, we cut the line into 4 groups of 13 cards. If we want the aces to be in each group, we have 13 places for each ace per group.

The result is then $\dfrac{13^4}{52 \choose 4}=\dfrac{28'561}{270'725}=\boxed{\dfrac{2'197}{20'825}}$ .