Question

I am playing a game of cards in which 52 distinct cards are allocated randomly
to four players (one of whom is me), each player receiving 13 cards. Four of the
cards are aces; one of these is called the ace of spades and another is called the ace
of clubs.

(i) What is the probability that I receive neither the ace of spades nor the ace
of clubs?

(ii) What is the probability that I receive at least one ace?

Answer 1

ok last one is easy enough. the probability you get at least one ace is 1 - probabiliy that you get no aces. that is
\[\frac{\dbinom{48}{13}}{\dbinom{52}{13}}\]

Answer 2

i think first one is similar. if you are not allowing either of the two aces, you have 50 cards as possibilites, so it should be
\[\frac{\dbinom{50}{13}}{\dbinom{52}{13}}\]

Answer 3

do you know how to compute these numbers?

Answer 4

Yes, that seems to make sense. Thanks a lot!

Answer 5

don't forget the first answer should be 1 - what i wrote

Answer 6

do you have the answer? we can check them because these numbers are easy to compute

Answer 7

Yes, the correct answer for (i) was 19/34, which is what you said. Unfortunately I don't have the answer for the second one, but the method makes sense. Cheers!

Answer 8

ok good.

Answer 9

right first one is
\[\frac{39\times 38}{52\times 51}=\frac{19}{34}\]