Is there an alternative intuition for solving the probability of having one ace card in every bridge player's hand?

Solution 1:

An alternative intuition is to see it as 4 groups of 13 with 4 friends all in different groups.

The first can be in any group, the 2nd has 39 permissible slots out of 51, and so on.

Thus Pr = $\frac{39}{51}\cdot\frac{26}{50}\cdot\frac{13}{49}$

Solution 2:

The solution is arguably simpler if you distinguish between permutations within the hands. There are $4!$ ways of distributing the aces over the $4$ players, and $13^4$ possible positions for the aces in the hands. That leaves $48!$ possibilities for filling the remaining $48$ slots. The size of the sample space in this perspective is just $52!$, so the desired probability is

$$\frac{4!\cdot13^4\cdot48!}{52!}\;.$$

Solution 3:

Rather than trying to rephrase, I think a great choice is the explanation by F. Scholz here: