Question

Assume that 5 cards are dealt at random from a standard deck of 52 cards (there are 4 suits in the deck and 13 different values (ranks) per each suit). We refer to these 5 cards as a hand in the rest of this problem. Calculate the probability of each of the following events when dealing a 5-card hand at random. (a) Exactly one pair: This occurs when the cards have numeric values a, a, b, c, d, where a, b, c, and d are all distinct. (b) Exactly two pairs: This occurs when the cards have numeric values a, a, b, b, c, where a, b, and c are all distinct. (c) Only three of a kind: This occurs when the cards have numeric values a, a, a, b, c, where a, b, and c are all distinct. (d) Four of a kind: This occurs when the cards have numeric values a, a, a, a, b (clearly, b must be different from a because there are only 4 suits in the deck). (e) Full house: This occurs when the cards have numeric values a, a, a, b, b, where a and b are distinct. (f) Any of the scenarios above will lead to having at least a pair in the hand, and having at least a pair in the hand implies one of the events above must be true. Now, use the probabilities calculated in parts (a)–(e) to calculate the probability that we see at least a pair in the hand. Your answer has to be exactly 49.29%, ignoring rounding error.

Answer 1

In this problem one has to know the concept of combination

that is the formula

Also here we have considered combination as we are not bothering about the order of the chosen cards.

We are just asked to select/choose cards . Hence here we are not considering arrangement of chosen cards amongst themselves.