What is the probability that four players who each receive ten cards together receive less than four aces? A deck of cards contain 52 cards, including 4 aces. Suppose each player gets 10 cards and the other 12 cards are kept aside, what is the probability that the four players together have less than four aces?
My answer:


*

*N = Total possibilities

*n = possibilities where the four players together have four aces 

*p = probability that four players together have less than four aces =
$\frac{N-n}{N}$


$$N = \frac{52!}{10!10!10!10!12!}=9.71*10^{32}$$
For n, I consider the possibilities of the distribution of the four aces among the four players:


*

*Each player gets 1 ace

*One player gets 2 aces, two players get 1 ace, one player gets 0 aces

*Two players get 2 aces, two players get 0 aces

*One player gets 3 aces, one player gets 1 ace, two players get 0 aces

*One player gets 4 aces, three players get 0 aces


$$n = \frac{4!}{1!1!1!1!}*\frac{36!}{9!9!9!9!}+\frac{4!}{2!1!1!0!}*\frac{36!}{8!9!9!10!}+\frac{4!}{2!2!1!0!}*\frac{36!}{8!8!9!10!}+\frac{4!}{3!1!0!0!}*\frac{36!}{7!9!10!10!}+\frac{4!}{4!0!0!0!}*\frac{36!}{6!10!10!10!}$$
(36 is used because we are observing the set of cards excluding the 4 aces and the 12 cards kept aside)
Using the values of N and n, p can be calculated. Is this correct? If so, is there a more efficient way to count the possibilities for n? Thanks!
 A: You do not need to consider the player separately.
You can consider the problem as taking $40$ cards from the deck of $52$ and calculate the probability that one or more aces are missing.
And if they are missing, they are in the remaining $12$ cards. So better work on finding the probability of having one ace or more in the $12$ cards = $1$ - prob. $0$ aces.
A: The players collectively receive $40$ cards.  If they collectively receive all four aces, then they receive $36$ of the other $48$ cards in the deck.  Hence, the probability that the players collectively receive all four aces is 
$$\frac{\dbinom{4}{4}\dbinom{48}{36}}{\dbinom{52}{40}}$$
Hence, the probability that the players collectively receive fewer than four aces is 
$$1 - \frac{\dbinom{4}{4}\dbinom{48}{36}}{\dbinom{52}{40}}$$
A: $\underline{Consider\; only\; the\; aces\;distributed}$
$40$ out of $52$ slots are allotted for players,
P(All the aces are with players) $ = \frac{40}{52}\frac{39}{51}\frac{38}{50}\frac{37}{49}$
Take its complement to get the answer
A: You can look at the way in which the aces can be in the complement. If there is at least one ace in the remaining 12 cards, then the 4 players will have fewer than 4 aces. The probability of at least one ace is in complement is 1-P(no aces in complement), which is the probability of selecting 12 cards none of which is an ace:
$$\frac{48 \choose 12}{52 \choose 12} = \frac{\frac{48!}{12!36!}}{\frac{52!}{12!40!}} = \frac{1406}{4165}$$
The desired probability is thus:
$$1 - \frac{1406}{4165} = \frac{2759}{4165} \approx 0.6625$$
