Probability need help A poker hand consists of 5 randomly de

A poker hand consists of 5 randomly deal cards. Compute the probability of Getting 3 of a kind (but not a full house) Getting a full house (3 of a kind and 2 of a kind)

Solution

a)

There are 13 ranks to choose from for the three of a kind and 4 ways to arrange 3 cards among the four to choose from. There are (12 choose 2) = 66 ways to arrange the other two ranks to choose from for the other two cards. In each of the two ranks there are four cards to choose from. Thus there are 13 * 4 * 66 * 4² = 54,912 ways to form a three of a kind. [Alternatively, the last two cards can be chosen in 48 * 44 / 2 ways. The fourth card must have a different rank to the three of a kind giving 52 - 4 = 48 possibilities, and the fifth card must be another different rank giving 52 - 4 = 48 possibilities. The order of the last two card is irrelevant so we must divide by 2.]

b)

The 3 of a kind can be chosen in 13 * (4 choose 3) ways as there are 13 possible ranks and ( 4 choose 3) ways to choose 3 cards from a given rank. The pair (or 2 of a kind) can be chosen in 12 * (4 choose 2) ways as there are 12 = 13 - 1 possible different ranks and (4 choose 2) ways to choose 2 cards from a given rank. In total, there are 13 * 4 * 12 * 6 = 3,744 ways to form a full house.