2 A poker hand is a set of 5 cards randomly chosen from a deck of 52 cards. Find the probability of a (a) royal flush (ten, jack, queen, king, ace in a single suit). (b) straight flush (five in a sequence in a single suit, but not a royal flush). (c) four of a kind (four cards of the same face value). (d) full house (one pair and one triple, each of the same face value). (e) flush (five cards in a single suit but not a straight or royal flush). (f) straight (five cards in a sequence, not all the same suit). (Note that in straights, an ace counts high or low.)
Solution:
The number of possible hands you can achieve are \(\begin{pmatrix} 52 \\ 5 \end{pmatrix} = (\frac { 52! }{ 5!(52-5)! } ) = 2,598,960\)
There are 4 possible way to get a royal flush, one for each suit. So the probability of a royal flush is \(\frac { 4 }{ 2,598,960 } = 0.000001539%\).
Since we are not including a royal flush, the highest card we look at is a king and the lowest card we look at is an ace. There are 9 hands we can make for our 5 card hand, from a 5 high straight flush to a king high straight flush. There are 4 possible suits for each of these 9 hands so the probability of getting a straight flush is \(\frac { 9*4 }{ 2,598,960 } = \frac { 36 }{ 2,598,960 } = 0.00001385%\).
There are 13 possible ways to get a four of a kind, since there are 13 different kinds of cards in a deck. We have 5 cards in a poker hand so the remaining card will be from the 48 remaining cards. So the probability of getting a four of a kind is \(\frac { 13*48 }{ 2,598,960 } = \frac { 624 }{ 2,598,960 } = 0.0002401%\).
We need a three of a kind and a pair to form a full house. For the three of a kind we have 13 ways to get a three of a kind, since there are 13 different kind of cards. The three of a kind can be formed from any 4 suits. The last two cards in the 5 card hand is a pair, and there are 12 ways to get a pair since one kind of card is used for the three of a kind. The suit can be any 2 of 4 suits since the pair is two cards that have the same face value, so there are \(\begin{pmatrix} 4 \\ 2 \end{pmatrix} = (\frac { 4! }{ 2!(4-2)! } ) = 6\) possible ways to get 2 our of 4 suits. So the probability of getting a full house is \(\frac { 13*4*12*6 }{ 2,598,960 } = \frac { 3744 }{ 2,598,960 } = 0.001441%\).
To get a flush we need all 5 cards in the hand to have the same suit. There are 13 different kinds of cards and only 5 cards in the hand, so \(\begin{pmatrix} 13 \\ 5 \end{pmatrix} = (\frac { 13! }{ 5!(13-5)! } ) = 1278\). We are told not to include any straights, so there are 10 straights that can be formed from king high straight to 5 high straight to take out. 1287 - 10 = 1277. There are 4 suits, so the probability of getting a flush is \(\frac { 1277*4 }{ 2,598,960 } = \frac { 5108 }{ 2,598,960 } = 0.001965%\).
To get a straight we need 5 consecutive numbers, the suit of the card is not important. So there are 10 possible ways to a straight. Once we get the first card we would already know the 4 other cards that follow, we just need to know what suit the cards are. So there are 4 suits to be determined for the 5 cards, so 4 * 4 * 4 * 4 * 4 or = 1024. There are 10 possible ways to get a straight, so 10 * 1024 = 10240. We also need to subtract the number of straight flushes and royal flushes, so the probability of getting a straight is \(\frac { 10240-36-4 }{ 2,598,960 } = \frac { 10200 }{ 2,598,960 } = 0.003925%\).