Originally Posted by

**awkward** I had difficulty seeing the "book solution", but here is, I think, a plausible explanation.

Suppose one of the players is Fred, and let's see if we can find the probability that he gets all 4 aces. For convenience of analysis, let's say the first 13 cards in the deck are dealt to Fred. If we simply distinguish between aces and non-aces, and we consider the aces indistinguishable, then there are $\displaystyle \binom{52}{4}$ possible locations of the aces in the deck, all of which we assume are equally likely. Of these, $\displaystyle \binom{13}{4}$ have all the aces among the first 13 cards. So Fred's probability of getting all the aces is

$\displaystyle \frac{\binom{13}{4}}{\binom{52}{4}}$.

Multiply by 4 to obtain the probability that some player, not necessarily Fred, gets all the aces.