A poker deck consists of cards ranked 2; 3; 4; 5; 6; 7; 8; 9; 10; J; Q;K;A (13 d

Question

A poker deck consists of cards ranked 2; 3; 4; 5; 6; 7; 8; 9; 10; J; Q;K;A (13 different ranks), each in four suits, for a total of 52 distinct cards.
(a) What is the probability that a five-card poker hand drawn from a poker deck consists only of cards ranked 8; 9; 10; J; Q;K;A?
(b) Find a probability of Three of a kind. This is, three cards of the same denomination and two cards from two other denominations.
(c) Find a probability of Two pairs. That is, two cards of one denomination, two cards of a different denomination and one card from yet another denomination.

Explanation / Answer

a. There are 4 possibilities of Royal flush that is of club, diamond, heart and spade. There are nine sequences in each suit, including the Royal flush, therefore, if the Royal flush is excluded, there remains 8 sequences in each suit. This results into 32 possible straight flushes. Now, consider the given sequence, since each card can be of any suit, there are (1C4)^5 possible straights. That is there will be 9*4^5-32 straight flush-4 royal flush, that is 9180 possibilities. Note, there are 9 possible five card sequences. Therefore, possibility of straight is: 9180/52C5=0.004

b. The rank of three cards can be chosen out of 13 possibilities in 13C1 ways, and then out of 4 suits availble, the suit of the cards can be chosen in 4C3 ways. Since, the remaining 2 cards has to be from different denominations, it can be chosen in 12C2 ways. Each of these two cards can be of any of the four suits therefore, the possibilities of getting Three of a kind is 13C1*4C3*12C2*(1C4)^2=54912.

P(three of a kind)=54912/52C5=54912/2598960=0.02

c. Two cards of one rank can be chosen in 13C2 ways, followed by 4C2 ways of choosing the suit, the two cards of another rank can be chosen in 4C2 ways, and the remaining card can be chosen in 44C1 ways. Therefore, total possibilities is: 13C2*4C2*4C2*44C1=123552.

P(two pairs)=123552/52C5=123552/2598960=0.05

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