Calculate the probability of the following poker hands for a single hand of five
ID: 3040819 • Letter: C
Question
Calculate the probability of the following poker hands for a single hand of five-card stud poker. Fully justify your answers. For those that don’t know, this means you shuffle the deck and deal exactly five cards. So, you need not worry about cards being dealt to other players or to the dealer. For clarity, we will assume that ‘wrap-around’ straights are not allowed and ace is the high card. This is the type of straight that includes an ace, a king, and a two, so acceptable straights are 2-6 through 10-Ace.
Hand A: a royal flush in spades (a royal flush is composed of the ace, king, queen, jack, and 10 of a single suit).
Hand B: a royal flush in any suit
Hand C: a straight in hearts
Hand D: a straight in any suit
Explanation / Answer
We must understand that the procedure of a five-card stud poker is drawing 5 cards simultaneuosly from a full deck of cards, which is the EQUIVALENT of drawing 5 cards one-by-one from a full deck of cards WITHOUT replacement
Each draw is DEPENDENT on previous draws as there is NO REPLACEMENT OF DRAWN CARD
-- Hand A --
royal flush: consists of the ace, king, queen, jack and ten of a suit, here, we want spades
=> P(royal flush in spades) = Probability of getting those 5 cards drawn in ANY ORDER
= (Number of ways these 5 can be arranged in a series)*(Probability of any 1 such series arising in five-stud poker)
Number of ways 5 DISTINCT objects can be arranged in a series = 5! = 5*4*3*2*1 = 120
=> Number of ways these 5 can be arranged in a series = 5! = 120
Let us say 1 sample series of these 5 objects be: ace first, king second, queen third, jack fourth, ten fifth
P(this series to occur) = P(ace first) * P(king second) * P(queen third) * P(jack fourth) * P(ten fifth) (BECAUSE all are DEPENDENT events, we can multiply by changing DENOMINATOR (i.e., total cards remaining))
=> P(this series to occur) = (1/52)*(1/51)*(1/50)*(1/49)*(1/48)
=> P(royal flush in spades) = (5!)*[ (1/52)*(1/51)*(1/50)*(1/49)*(1/48) ] = 0.000000384769292 [ANSWER A]
-- Hand B --
royal flush in any suit: consists of the ace, king, queen, jack and ten of a suit, here, we may have ANY suit
=> P(royal flush in any suit) = Probability of getting those 5 cards of a single suit drawn in ANY ORDER
As the calculation in Hand A solution was independent of the actual suit being targeted upon, we can say that the probability of a royal flush in a suit will be equal to that in any other suit and hence, the total probability of a royal flush in any suit is the sum of probability of a royal flush in each suit
=> P(royal flush in any suit) = (Number of suits)*(Number of ways these 5 can be arranged in a series)*(Probability of any 1 such series arising in five-stud poker)
Number of suits = 4
Number of ways 5 DISTINCT objects can be arranged in a series = 5! = 5*4*3*2*1 = 120
Let us say 1 sample series of these 5 objects be: ace first, king second, queen third, jack fourth, ten fifth
P(this series to occur) = P(ace first) * P(king second) * P(queen third) * P(jack fourth) * P(ten fifth) (BECAUSE all are DEPENDENT events, we can multiply by changing DENOMINATOR (i.e., total cards remaining))
=> P(this series to occur) = (1/52)*(1/51)*(1/50)*(1/49)*(1/48)
=> P(royal flush in any suit) = 4*(5!)*[ (1/52)*(1/51)*(1/50)*(1/49)*(1/48) ] = 0.00000153907 [ANSWER B]
-- Hand C --
A straight in any suit: Series of 5 consecutive numbers of a single suit coming up, in any order
P(A straight in hearts) = (Number of Possible straights combinations of Hearts)*(Number of ways these 5 can be arranged in a series)*(P(1 such series of Hearts to occur))
As wrap-around straights are not allowed, only straights from 2-6 to 10-Ace are allowed => 9 ways!
=> Number of Possible straights combinations of Hearts = 9
Number of ways 5 DISTINCT objects can be arranged in a series = 5! = 5*4*3*2*1 = 120
An example of a straight: (2-6) => {2,3,4,5,6}
=> Number of ways these 5 can be arranged in a series = 5! = 120
P(1 such series of Hearts to occur) = P({2,3,4,5,6} of Hearts to occur)
= P(2 first) * P(3 second) * P(4 third) * P(5 fourth) * P(6 fifth)
(BECAUSE all are DEPENDENT events, we can multiply by changing DENOMINATOR (i.e., total cards remaining))
=> P({2,3,4,5,6} of Hearts to occur) = (1/52)*(1/51)*(1/50)*(1/49)*(1/48)
Hence, P(A straight in hearts) = 9*(5!)*[ (1/52)*(1/51)*(1/50)*(1/49)*(1/48) ] = 0.00000346292 [ANSWER C]
-- Hand D --
A straight in any suit: Series of 5 consecutive numbers of a single suit coming up, in any order
As the calculation in Hand C solution was independent of the actual suit being targeted upon, we can say that the probability of a straight in a suit will be equal to that in any other suit and hence, the total probability of a straight in any suit is the sum of probability of a straight in each suit
P(A straight in any suit) = (Number of suits)*(Number of Possible straights combinations in a single suit)*(Number of ways these 5 can be arranged in a series)*(P(1 such series in a single suit to occur))
As wrap-around straights are not allowed, only straights from 2-6 to 10-Ace are allowed => 9 ways!
=> Number of Possible straights combinations in a single suit = 9
Number of ways 5 DISTINCT objects can be arranged in a series = 5! = 5*4*3*2*1 = 120
An example of a straight: (2-6) => {2,3,4,5,6}
=> Number of ways these 5 can be arranged in a series = 5! = 120
P(1 such series in a single suit to occur) = P({2,3,4,5,6} in a single suit to occur)
= P(2 first) * P(3 second) * P(4 third) * P(5 fourth) * P(6 fifth)
(BECAUSE all are DEPENDENT events, we can multiply by changing DENOMINATOR (i.e., total cards remaining))
=> P({2,3,4,5,6} in a single suit to occur) = (1/52)*(1/51)*(1/50)*(1/49)*(1/48)
Hence, P(A straight in a single suit) = 4*9*(5!)*[(1/52)*(1/51)*(1/50)*(1/49)*(1/48)] = 0.00001385169 [ANSWER D]
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.