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!
Best Answer
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:
$$1 - \frac{1406}{4165} = \frac{2759}{4165} \approx 0.6625$$