Probability of Poker Hands

1 Probability of Poker HandsDrew 1, 2006In a standard deck of cards, there are 4 possible suits (clubs, diamonds, hearts, spades),and 13 possible values (2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, Ace). LetA, J, Q, Krepresent Ace, Jack, Queen and King, respectively. Every card has a suit and value , and everycombination is possible. Hence a standard deck contains 13 4 = 52 Poker hand consists of 5 unordered cards from a standard deck of 52. There are(525)= 2,598,9604 possible Poker Hands . Below, we calculate the Probability of each of thestandard kinds of Poker hand consists of values 10, J, Q, K, A, all of the same suit.

2 Since the valuesare fixed, we only need to choose the suit, and there are(41)= 4 ways to do straight flush consists of five cards with values in a row, allof the same may be considered as high or low, but not both. (For example,A,2,3,4,5 is a straight,butQ, K, A,2,3 is not a straight.) The lowest value in the straight may beA,2,3,4,5,6,7,8or 9. (Note that a straight flush beginning with 10 is a royal flush, and we don t want to countthose.) So there are 9 choices for the card values, and then(41)= 4 choices for the suit, givinga total of 9 4 = straight consists of five values in a row,notall of the same suit.

3 The lowest valuein the straight could beA,2,3,4,5,6,7,8,9 or 10, giving 10 choices for the card values. Thenthere are(41)5= 45ways to choose the suits of the five cards, for a total of 10 45= 10,240choices. But this value also includes the straight flushes and royal flushes which we do not wantto include. Subtracting the 40 straight and royal flushes, weget 10,240 40 = 10, flush consists of five cards, all of the same suit. There are(41)= 4 ways to choosethe suit, then given that there are 13 cards of that suit, there are(135)ways to choose the hand,giving a total of 4 (135)= 5,148 flushes.

4 But note that this includes the straight and royal flushes,which we don t want to include. Subtracting 40, we get a grandtotal of 5,148 40 = 5, of a hand consists of four cards of one value , and a fifth card of a differentvalue. There are(131)= 13 ways to choose the value for the quadruple. Then, among the cards1of this value , there are(44)= 1 ways to choose the quadruple. After this, there are(121)= 12ways to choose a value for the single from the remaining values, and(41)= 4 ways to choosethe single from the four cards of this value , for a grand totalof(131)(44)(121)(41)= 13 1 12 4= hand consists of three cards of one value , and two cards of a different are(131)ways to choose a value for the triple, then(43)ways to choose the triple fromthe four cards of this value .

5 Then, there are(121)ways to choose the value of the double fromthe remaining values, and(42)ways to choose the double from the four cards of this value , fora grand total of(131)(43)(121)(42)= 13 4 12 6= 3, of a hand consists of three cards of one value , and two more cards, each ofdifferent values. There are(131)ways to choose the value for the triple, and(43)ways to choosethe triple from the four cards of this value . Then there are(122)ways to choose two (unordered)values for the remaining singles, and(41)(41)to choose the singles from their respective values,for a grand total of(131)(43)(122)(41)(41)= 13 4 66 4 4= 54, hand consists of two pairs of different values, and a fifthcard of anotherdifferent value .

6 There are(132)ways to choose two (unordered) values for the two pairs, then(42)(42)to choose the pairs from the cards of these values. Then thereare(111)ways to choosea remaining value for the single, and(41)ways to choose the single from the four cards of thisvalue, for a grand total of(132)(42)(42)(111)(41)= 78 6 6 11 4= 123, hand consists of a pair of one value , and three additional cards, each ofdifferent value . There are(131)ways to choose a value for the pair, then(42)ways to choose the2pair from the four cards of this value .

7 Then there are(123)ways to choose three (unordered)values for the remaining three singles, and(41)3to choose suits for the singles, for a grand totalof(131)(42)(123)(41)3= 13 6 220 43= 1,098, all of this together, we obtain the following ranking of Poker Hands : Poker HandNumber of Ways to Get ThisProbability of This HandRoyal of a House3, , , of a Kind54, Pairs123, Pair1,098, ,302, , how did I compute the Probability of getting Nothing ?How would you answer the question: What is the Probability of getting Three of a Kindor better?

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