More Probability: Poker Hands and some issues in Counting
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1 More Probability: Poker Hands and some issues in Counting
2 Data From Thursday Everybody flipped a pair of coins and recorded how many times they got two heads, two tails, or one of each. We saw that the probability of getting two heads is.25 and that the probability of getting one of each is.50. The summary of the data is as follows. 2
3 Outcome Two Heads Two Tails One of Each Number Fraction
4 Everybody also rolled two dice 20 times and determined the sum. The tabulation of the data is as follows. 4
5 Sum of Two Dice Sum Number Fraction
6 The theoretical probability of getting a sum of 12 (e.g., rolling a double six) is 1/36, since of the 36 outcomes of rolling two dice, one of them results in a sum of 12. In decimal form, this is about.028, or very close to what the class data was. 6
7 Poker Hands 7
8 Poker is played with a deck of 52 cards. Each card has a suit and a value. 8
9 The suits are spades hearts diamonds and clubs. 9
10 There are 13 values: 2 through 10, Jack, Queen, King, and Ace. A poker hand is made up of 5 cards. The different poker hands are, from best to worst, are: 10
11 Royal Flush Straight Flush Four of a Kind 11
12 Full House Flush Straight 12
13 3 of a kind 2 pair 1 pair 13
14 Hand Example Royal Flush Straight Flush A, K, Q, J, 10 J, 10, 9, 8, 7 4 of a kind 8, 8, 8, 8, J Full House Flush Straight 4, 4, 4, J, J K, 10, 8, 7, 3 Q, J, 10, 9, 8 3 of a kind 8, 8, 8, J, 6 2 pair 8, 8, J, J, A 1 pair 8, 8, 10, J, K 14
15 Royal flush: A, K, Q, J, 10 in the same suit. Straight flush: 5 consecutive cards in the same suit. 4 of a kind: 4 cards of the same value. Full house: 3 cards of the same value and a pair. 15
16 Flush: 5 cards in the same suit. Straight: 5 consecutive cards (of any suit). An ace can be low or high. 3 of a kind: 3 cards of the same value. 1 pair: 2 cards of the same value. 16
17 How many poker hands are there? The number is how many ways you can choose 5 cards out of 52. This number is usually denoted 52C5, or ( 52 52C5, or 5 ). It is also called a binomial coefficient. 17
18 How many ways are there to choose 1 item out of 5? There are 5 ways. How many ways are there to choose 2 items out of 4? If the items are labeled a, b, c, d, we could have {a,b}, {a, c}, {a, d}, {b, c}, {b, d}, {c, d}, so there are 6 ways, provided that the order in which we choose them does not matter (like in a poker hand). 18
19 There is a formula to compute binomial coefficients for any pair of numbers. It is n! ncr = = r! * (n-r)! n*(n-1)*... *(n-r+1) r*(r-1)*... *2*1 where n! means 1 * 2 * 3 *... * n 19
20 Many scientific calculators have n! buttons, and many have buttons, or menu items to calculate ncr. Some values of n! are: 0! = 1 1! = 1 2! = 2 3! = 6 4! = 24 20
21 As an example of this formula, we see that nc1 = n for any number. What this means is that if you choose 1 item out of n, there are n ways to do this. Another example is that nc2 = n*(n-1)/2. This comes from canceling terms in the formula. 21
22 For example, 4C2 = 4! (2! * 2!) = 24 (2*2) = 6 Alternatively, 4C2 = 4*3/2 = 6. 22
23 If you calculate 52C5, you will get 52C5 = 2,598,960. Thus, there are 2,598,960 possible 5 card poker hands. Excel can compute ncr and n! It uses the commands =combin(n,r) and =fact(n), respectively. 23
24 What is the probability of getting a royal flush when dealt 5 cards? There are 4 ways to get a royal flush, since the only choice is which suit you get. So, the probability is which is about 1 out of 600,000 deals. 24
25 In order to compute the probability of other hands, one approach is to decide how many things you need to choose in order to write down a hand, and then determine how many ways each of the choices can occur. For example, to count royal flushes, we saw that you only have to choose a suit. 25
26 Counting Independent Events If one event does not affect the outcome of another, they are called independent. To count the number of ways a pair of independent events can occur, multiply the number of ways each way can occur. 26
27 Example Rolling two dice is an example of two independent events: what you get on one die does not affect what can happen on the other. Since there are 6 outcomes for rolling one die, there are 6 * 6 = 36 outcomes for rolling two dice. 27
28 How many ways are there to get a straight flush? To choose a given straight flush, you must choose a suit, and a starting (or ending) value for the 5 in a row. There are 4 choices for the suit. What suit you choose does not affect the choice of starting value for the 5 in a row. 28
29 There are 10 possible starting values (A through 10). However, if we want a straight flush which is not a royal flush, we cannot start at 10, so there are 9 choices. Therefore, there are 4 * 9 = 36 total straight flushes. 29
30 The probability of a straight flush (which is not a royal flush) is then
31 What is the probability of a 4 of a kind? An example is 8, 8, 8, 8, J. We must count how many 4 of a kinds there are. To get a 4 of a kind, you must choose the value of the 4 of a kind, and choose the remaining card. There are 13 choices for the value of the 4 of a kind. The 5th card can be any of the remaining 48 cards. So there are 48 choices for it. 31
32 The number of four of a kinds is then 13 * 48 = 624. and so the probability of a full house is
33 A full house consists of a 3 of a kind and a 2 of a kind. What is the probability of getting a full house? An example is 4, 4, 4, J, J. To have a full house you must choose the value of a 3 of a kind and the value of a 2 of a kind. You must also choose which 3 cards make up the 3 of a kind and which 2 make up the 2 of a kind. This is one of the more complicated counts. 33
34 There are 13C1 = 13 ways to choose the value of the 3 of a kind. There are then 12C1 = 12 ways to choose the value of the pair. There are 4C3 = 4 ways to choose the three cards for the 3 of a kind. There are 4C2 = 6 ways to choose the 2 cards for the pair. 34
35 So, the number of ways to get a full house is 13 * 12 * 4 * 6 = The probability of a full house is then a little worse than 1 out of 4000 hands. 35
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