Recall that an ordinary deck of cards has 52 cards uniquely identified by 1 of 4 suits (hearts=H, clubs C, diamonds D, spades=S) and by 1 of 13 face values or denominations (ace=A, 2,3,4,5,6,7,8,9,10, jack=.J, queen Q, king-K). A hand consists of 5 cards chosen from the 52.
A hand containing exactly one pair consisting of two cards with the same denomination but different suits and one triple consisting of three cards with the same denomination but different suits is called a full house. For example, a hand containing (4H, 4S, 3C, 3H, 3D) is a full house. The 4 of hearts and the 4 of spades is the pair in the this example and the 3 of clubs, 3 of hearts, and 3 of diamonds is the triple in this example.
Compute the probability of a hand containing a full house.
HINT: The denominator should be (5) =2,598,960.
(2)
Further, to compute the numerator, one should proceed by the multiplication rule. The process of forming exactly one pair consists of 4 steps (in my solution), so one must count the number of ways to perform each step and multiply them by the hint to use the multiplication rule.