Poker CombinationsDate: 05/09/97 at 08:04:55 From: David Fried Subject: Re: High School Precal (probability) Dr. Math, In a standard deck of cards, how many different ways are there for getting a straight, a flush, a straight flush, one pair, or two pairs? A straight consists of five cards in a row: (4,5,6,7,8) (A,2,3,4,5), (10,J,Q,K,A). Note: The "A" cannot be in the middle. A flush consists of five cards of the same suit: 5 clubs, 5 hearts, and so on. A straight flush consists of five cards of the same suit, in a row: (4,5,6,7,8), all hearts, for example. Two pairs consists of two separate pairs of two cards: 2 K's and 2 A's or 2 4's and 2 10's One pair consists of one pair of two cards. Could you give the proofs please? Thank you! Date: 05/14/97 at 08:09:06 From: Doctor Anthony Subject: Re: High School Precal (probability) How many different ways can you get: (1) A straight - i.e., 4, 5, 6, 7, 8, (mixed suits). The number of consecutive groups of 5 cards is 10. i.e., We can start with A, 2, 3, ....., 10. In each sequence of 5 we can choose each card in 4 ways (Clubs, Hearts,...), this gives 4^5 ways = 1024. So the total is 1024 x 10 = 10240. However, this total will include the straight flush, so we must subtract the number that are straight flushes. For the straight flush, the suit can be chosen in 4 ways. So the number of straight flushes is 10 x 4 = 40. The number of straights is 10240 - 40 = 10200. (2) Flush (i.e. 5 cards of same suit). You can choose the suit in 4 ways, and you can choose 5 cards from 13 in 13_C_5 = 1287 ways. The total number of ways of getting a flush is 4 x 1287 = 5148. Again we must subtract from this the number of ways of getting a straight flush. The number of flushes is 5148 - 40 = 5108. (3) Straight flush. As we have seen, there are 40. (4) One pair (i.e., 2 K's, or 2 6's etc.). Suppose we have 2 K's. The K's can be chosen in 4_C_2 ways = 6 ways The face value of the pair can be chosen in 13 ways. Now we need to choose 3 more cards, none of which match each other. So we choose 3 cards out of the 12 remaining values, which is 12_C_3 = 220 ways. But also each card can be 1 of the 4 suits, and so there are 4*4*4 ways to choose the suits. Total number of ways is 6 x 13 x 220 x 4 x 4 x 4 = 1,098,240. (5) Two pairs (i.e., 2 J's and 2 8's). We can choose the 2 J's in 4_C_2 ways = 6 ways We can choose the 2 8's in 4_C_2 ways = 6 ways We can choose the face value of the pairs in 13_C_2 = 78 ways We can choose the face value of fifth card in 11 ways, in each of the four suits, for 44 ways. Total number of ways is 6 x 6 x 78 x 11 x 4 = 123,552 -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/