Question

Consider five-card-hands out of a deck of 52 cards.

a) (10%) What is the probability that a hand selected at random contains five cards of the same suit? (The suits are hearts, spades, diamonds and clubs.) Show a formula and compute a final answer. Show your intermediate computations.

b) (10%) In how many different ways can a hand contain no more than 2 cards of the same kind. (E.g. not more than 2 queens.) Show a formula but you do not have to compute a final answer.

Answer 1

A hand has 5 cards out of a deck of 52 cards.

(a) For the number of hands with cards of the same suit, calculate as following. The first card can be any card. Hence, there are ways of choosing the first card.

The next four cards must be from the same suit as the first card. Note that after picking each card, the number of remaining cards of that suit decreases by 1.

Hence, the number of ways of choosing the next four cards is .

Using the multiplication principle, the final answer is .

(b) There are two cases to consider, when no more than 2 cards should be of the same kind.

1. There are no cards of the same kind. To count this, the first card can be anything. Hence 52 ways.
   
   The second card must not be of same kind as first. Hence, 4 less options are available. The number of ways is 48.
   
   Similarly, for the third, fourth and fifth card, there are 44, 40, and 36 possibilities, respectively.
   
   The total number of ways is thus .
   
2. There are two cards of the same kind. The first card can be anything. Hence 52 ways.
   
   Say the second card is the same kind as that of the first. There are 3 remaining cards of that kind. Hence, 3 ways to choose the second card.
   
   Now, all subsequent cards must have different kind. Hence, for the third card there are 48 possibilities. Similarly, for the fourth and the fifth, there are 44 and 40 possibilities, respectively.
   
   The total is .

Add both the cases to get the final answer as .

Comment in case of any doubts.