2008-12-03, 06:27 PM
I got this assignment today in data management that's due next week. Basically we have to figure out the probability of getting each type of hand with a poker hand of 5 cards.
I already finish them in class, but I just want to make sure they are all right.
So if any of the answers here are wrong, please tell me, but don't tell me the answer cause I'll figure it out myself.
Royal Flush: 4/52C5 => 0.00015390771%
Other Straight Flush: 9*4/52C5 => 0.00138516945%
Four of a Kind: 13*48/52C5 => 0.024009603%
Full House: (13*4C3)*(12*4C2)/52C5 => => 0.144057623%
Flush: (13C5*4-40)/52C5 => 0.196540154%
Straight: (10*4*4*4*4*4-40)/52C5 => 0.392464678%
Three of a kind: ((13*4C3)*(48*44)/2!)/52C5 => 2.112845138%
Two Pairs: ((13*4C2*12*4C2)/2!*44)/52C5 => 4.753901561%
One pair: ((13*4C2)*(48*44*40)/3!)/52C5=>42.25690276%
No pair 100% minus everything above. =>~50% (too lazy to calculate now)
I already finish them in class, but I just want to make sure they are all right.
So if any of the answers here are wrong, please tell me, but don't tell me the answer cause I'll figure it out myself.
Royal Flush: 4/52C5 => 0.00015390771%
Other Straight Flush: 9*4/52C5 => 0.00138516945%
Four of a Kind: 13*48/52C5 => 0.024009603%
Full House: (13*4C3)*(12*4C2)/52C5 => => 0.144057623%
Flush: (13C5*4-40)/52C5 => 0.196540154%
Straight: (10*4*4*4*4*4-40)/52C5 => 0.392464678%
Three of a kind: ((13*4C3)*(48*44)/2!)/52C5 => 2.112845138%
Two Pairs: ((13*4C2*12*4C2)/2!*44)/52C5 => 4.753901561%
One pair: ((13*4C2)*(48*44*40)/3!)/52C5=>42.25690276%
No pair 100% minus everything above. =>~50% (too lazy to calculate now)

