In a poker game, 5 cards are dealt... In a poker game, 5 cards are dealt from a

Question

In a poker game, 5 cards are dealt...

In a poker game, 5 cards are dealt from a standard 52 card deck that has been well shuffled. You are the only player in this scenario. How many possible 5 card hands are possible? What is the probability that you are dealt two pairs? What is the probability that you are dealt a three of a kind or 4 of a kind? What is the probability that you are dealt a full house? What is the probability that you are dealt a flush, three of a kind, or 4 of a kind? (any type of flush is acceptable

Explanation / Answer

a. In Poker game, 5 cards are randomly chosen from a deck of 52 cards. Therefore, possible number of 5 card hands is 52C5=52!/5!(52-5)!=2598960.

b. Two pairs means two cards of one rank, two cards of another rank and the remaining card is of another rank. Therefore, there are 13C2 ways of choosing rank of the two cards, followed by 4C2 ways of choosing the suits, each of these two cards can be any of the 4 suits, therefore, 4C2, and then there is 44C1 way of choosing the remaining card.

Therefore, probability of two pair=13C2*4C2*4C2*44C1/52C2=123552/2598960=0.05

c. Three of a kind is three cards of one rank and remaining two cards, each of different rank. Therefore, in 13C1 ways the rank of three card is chosen, then 4C3 ways of choosing suits of the three cards, followed by 12C2 ways of choosing the remaining cards. Again, each of this two cards can be of four suits, therefore, 4C1*4C1 ways.

Therefore, the required probability: 13C1*4C3*12C2*4C1*4C1/52C2=54912/2598960

Four of a kind means four cards of one rank, and remaining card of another rank. Therefore, rank of four cards can be chsen in 13C1 ways, and remaining 1 card is chosen in 48C1 ways. Therefore, required probability is :13C1*48C1/52C2=624/2598960

P(three of a kind or four of a kind)=P(three of a kind)+P(four of a kind)=54912/2598960+624/2598960=0.02

d. Full house means three cards of one rank, and two cards of another rank. Therefore, rank of three cards can be chosen in 13C1 ways, followed by 4C3 ways of chooisng the suits. The remaining two cards of another rank can be chosen in 12C1 ways, followed by 4C2 ways of choosing the suits. Therefore, required probability is:

13C1*4C3*12C1*4C2/52C2=3744/2598960=0.001

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