Question 5 Part B Solution

Counting: Combinations Question 5, Part B Solution

Part I: In his hand of five cards, Shinobi received one pair: a nine (9) of spades and a nine (9) of hearts. Calculate how many ways Shinobi could have gotten a “one pair”. ( Note: A “one pair” hand consists of 2 cards of the same face value and 3 other cards. You do not match the pair with anything else ).

For this question we will use the “choose” formula : or C Where: “ n ” is the number of objects to choose from. “ r ” is the number of objects to be arranged. It means that given a set of " n " objects, how many different groups of " n " objects can be chosen where order doesn't mater. Since the orders in which the objects are arranged don’t matter we call it a “combination." n! (n-r)!r! r n

Solving: 3 * * * OR our equation could be simplified to look like this: This is our equation: Using: ( ) 13 C 1 * 4 C 2 * 12 C 3 * 4 C 1 3 12 C 2 4 C 13 C 1 4 C 1 4 C 1 * 4 C 1 * C r n

Each of our choices has a different value: Choosing our face value for our one pair. There are thirteen face values to choose from and from that we need one face value for our pair. = = Choosing our suits for our one pair. There are four suits in a standard 52 card deck and we need two suits for our paired cards of the one face value. = We have three other cards that don't match with our one pair. Therefore we are choosing 3 more face values out of the remaining 12 (one face value is already used for our one pair). = Choosing the suits for the other 3 cards. Each of the three cards has four suits to choose from, but they only need to have one suit each. 4 C 1 * 4 C 1 * * 4 C 1 13 C 1 * 4 C 2 * 12 C 3 13 C 1 2 4 C 12 C 3 4 C 1

We then solve the equation: = 13 * 6 * 220 * 4 * 4 * 4 = 1098240 4 C 1 * 4 C 1 * * 4 C 1 13 C 1 * 2 4 C * 12 C 3 There are 1098240 ways that Shinobi could have gotten the one pair hand.

An alternate way to look at it: We can also solve the problem using the other choose formula: * * * 3 = 1098240 ways n! (n-r)!r! 4 C 1 * 4 C 1 * 4 C 1 13 C 1 2 4 C 12 C = 13! 4! 12! 4! * 12! * 2! 2! * 9! 3! * 3! 3! 4! 4! 3! * 4! * * 13! * 4! 12! (13-1)! 1! (4-2)! 2! (12-3)! 3! (4-1)! 1! (4-1)! 1! * 4! (4-1)! 1! * 4! =

Part II: Shinobi has five cards in his hands. Two of which are a nine (9) of hearts and a Queen of hearts. The other 3 cards had no value to him so he thought that he could get rid of them to obtain a flush in the hearts suit. If Shinobi discarded the 3 cards and kept the nine (9) of hearts and the Queen of hearts, how many ways can Shinobi obtain a flush?

Things to look for in the Question: Shinobi has five cards in his hands. Two of which are a nine (9) of hearts and a Queen of hearts . The other 3 cards had no value to him so he thought that he could get rid of them to obtain a flush in the hearts suit. If Shinobi discarded the 3 cards and kept the nine (9) of hearts and the Queen of hearts, how many ways can Shinobi obtain a flush? (Note: A flush consists of five (5) cards of the same suit, but are not in sequence. It does not include straights or a royal flush)

Solving: There are 10 straights in flush. Since we are looking for ways to just obtain flushes, we will want to exclude all the possible straights we can get in a flush. If we had the straights, then the hand wouldn't be called a flush. Therefore we have to subtract the number of straights to keep the hand strictly a flush as shown in the following slides. A 2 3 4 5 6 7 8 9 10 J Q K A

We will solve our problem using the same equation we did previously. Therefore, we would get this: r n C * C 11 3 1 C 1 - 10

We need 3 more face values to complete our flush because we have a nine (9) and a Queen already. Remember a flush consists of five cards. Since we already used two of the face values, we have 11 left to choose from. We want only one suit because a flush consists of five cards of different face values in the same suit. Our one (1) suit to choose from is hearts because that's the suit we are looking for and it can only be arranged one (1) way. 10 We are subtracting the number of straights in a flush. If we had the straights, then the hand wouldn't be called a flush. Therefore we have to subtract the number of straights to keep the hand strictly a flush. = = = What it means: * C 11 3 1 C 1 - 10 C 11 3 1 C 1

Finishing it off: = 165 * 1 - 10 = 155 * C 11 3 1 C 1 - 10 There are 155 ways for Shinobi to gain a flush when he already has the two cards: nine (9) of hearts and Queen of hearts.

An alternate way: 11! (11-3)! 3! 1! (1-1)! 1! - 10 * * Using: 11! 8! 3! 0! 1! = = 1! 155 ways n! (n-r)!r!

Part III: Shinobi has obtained his flush in the suit of hearts. His five cards are: three (3), five (5), nine (9), Jack, and Queen. Due to the constant bluffing of the two suspicious guests, the other players except for Shinobi have folded their hands. Shinobi absolutely does not want to give them even an ounce of his money. If Shinobi decides not to fold and keep his flush, how many ways can the others get a hand to beat his flush?

Things to look for in the in the question: Shinobi has obtained his flush in the suit of hearts. His five cards are: three (3), five (5), nine (9), Jack, and Queen. Due to the constant bluffing of the two suspicious guests, the other players except for Shinobi have folded their hands. Shinobi absolutely does not want to give them even an ounce of his money. If Shinobi decides not to fold and keep his flush, how many ways can the others get a hand to beat his flush? (Note: Suits have no value)

This is Shinobi's hand. A flush in the suit, hearts using the cards, two (2), five (5), nine (9), Jack, and Queen. Since we are looking for how many ways the other person can get a hand to beat Shinobi's flush, we have to calculate the ways we can get each hand. The hands with a higher value than the flush are the Full house, Four of a Kind, Straight Flush, and Royal flush.

Full house: A full house consists of 3 cards of one face value and 2 cards of another . For example: Our equation for this problem would look like this: 3 ( ) * C 1 4 C 8 + 3 C 3 5 C 1 * ( ) * 4 C 2 * 7 C 1 ( ) + 2 3 C * 5 C 1 ( )

Let's break the equation up so we can work with it. We have four suits to choose from and we only need three of them for our three cards in our full house. = = We have eight face values to choose from for those three cards minus the 2, 5, 9, Jack, and Queen because we've already used one suit (hearts) for those cards. = We have three suits to choose from minus hearts. Choosing the face values from 2, 5, 9, Jack, and Queen = Choosing our three cards of one face value... 3 C 3 5 C 1 * ( ) + ( ) * 3 4 C C 1 8 3 4 C C 1 8 3 C 3 5 C 1

* ( ) + * ( ) We have four suits to choose from and we only need two of them for our two other cards. = = = = The face values we can choose from for our two cards, minus the face values in our hand and the three previously chosen for the full house. Choosing the suit for the two cards minus hearts. Choosing a face value from 2, 9, 5, Jack, and Queen. Choosing our two cards of another face value... 4 C 2 7 C 1 2 3 C 5 C 1 4 C 2 7 C 1 2 3 C 5 C 1

We have a case restriction because we can't use five of the cards from the hearts suit. Those cards are in our hands. So we have to find the ways we can get the face values of the other suits and the remaining unused face values in hearts seperately. We add them together afterwards to get the total way to obtain a full house. We multiply the other values together because they can be arranged in different ways together as well. 3 ( ) * C 1 4 C 8 + 3 C 3 5 C 1 * ( ) * 4 C 2 * 7 C 1 ( ) + 2 3 C * 5 C 1 ( )

= (4 * 8) + (1 * 5) * (6 * 7) + (3 * 5) = 32 + 5 * 42 + 15 = 257 There are 257 ways for our opponent to get a Full house. 3 ( ) * C 1 4 C 8 + 3 C 3 5 C 1 * ( ) * 4 C 2 * 7 C 1 ( ) + 2 3 C * 5 C 1 ( )

We can also solve the problem using the other choose formula: = = (4 * 8) + (1 * 5) * (6 * 7) + (3 * 5) = 257 ways = 32 + 5 * 42 + 15 n! (n-r)!r! 3 ( ) * C 1 4 C 8 + 3 C 3 5 C 1 * ( ) * 4 C 2 * 7 C 1 ( ) + 2 3 C * 5 C 1 ( ) * 4! (4-3)! 3! 8! (8-1)! 1! + 3! (3-3)! 3! * 5! (5-1)! 1! 4! * (4-2)! 2! * 7! (7-1)! 1! 3! + (3-2)! 2! 5! * (5-1)! 1!

Four of a Kind A four of a kind consists of four cards of one face value and one other card . For Example: Our equation for this problem would look like this: 4 C 4 * 8 C 1 ( ) * 4 C 1 * 7 C 1 ( ) + ( ) 8 1 C 1 * C 1

Let's break the equation up so we can work with it. We have four suits in a standard deck of cards and we need to choose from all of them for our four of a kind . = We have eight face values to choose from minus the ones in our hand since the face values in our hand our from one suit. You need all four suits for a face value in this hand. = ( ) 4 C 4 * 8 C 1 4 C 4 8 C 1

( ) * = = We our choosing the suit for the one extra card. Since we only need the one card, we only choose the one suit out of four. We are choosing the face value for the card. This is minus the face values in our hand and for the face value we just chose for our four of a kind. 4 C 1 7 C 1 4 C 1 7 C 1

( ) * = = We are now figuring out how many cards we can get from the hearts suit. That is one object to choose and one way to arrange it. There are cards remaining to choose from the hearts suit since we've already obtain them in our hand. 1 C 1 8 C 1 1 C 1 8 C 1

= (1 * 8) * (4 * 7) + (1 * 8) = 232 Finishing our solution... There are 232 ways for the opponent to get a Four of a kind. ( ) 8 1 C 1 * C 1 + 4 C 1 * 7 C 1 ( ) * 4 C 4 * 8 C 1 ( )

An alternate way to solve the problem... = = 232 = (1 * 8) * (4 * 7) + (1 * 8) 4 C 4 * 8 C 1 ( ) * 4 C 1 * 7 C 1 ( ) + ( ) 8 1 C 1 * C 1 ( ) 4! (4-4)! 4! * 8! (8-1)! 1! * 4! (4-1)! 1! 7! * (7-1)! 1! ( ) + 1! ( ) (1-1)! 1! 8! * (8-1)! 1!

A straight flush consists of 5 cards in sequence and of the same suit . It does not include Ten, Jack, Queen, King, and Ace (that's royal flush). For example: Straight Flush Our equation is: 9 * 3 C 1

We've used these cards in our hand A 2 3 4 5 6 7 8 9 10 J Q K A We have three suits to choose from because in the hearts suit, we don't have any sequences since we've used up five of the cards. = Although we have ten cards in sequence (a straight), one of them is a Royal flush and we don't count that. So really there are nine possible sequences in a straight flush. = 9 1 C 3 A 2 3 4 5 6 7 8 9 10 J Q K A 9 * 3 C 1

= 9 * 3 = 27 An alternate way... = 9 * 3! (3-1)! 1! = 27 9 * 3 C 1 There are 27 ways for the opponent to get a straight flush. 9 * 3 C 1

A royal flush consists of Ten, Jack, Queen, King, and Ace all in the same suit. For example: Royal Flush Our equation is: 3 C 1 1 *

What does it all mean? 1 = = There is only one way to get a 10, Jack, Queen, King, and Ace in a sequence. We have only three suits to choose from, spades, clubs, and diamonds because we've already used the Jack and Queen of hearts in our hand already. And from those three suits, we only need to choose one for our Royal flush. 3 C 1 1 * 3 C 1

Finishing our question off... An alternate way... There are 3 ways for the opponent to get a Royal Flush. 3 C 1 1 * 1 * = 9 = 3 1 * = 3! (3-1)! 1! = 3 3 C 1 1 *

Lastly... To finish off and give the actual answer to the question, we must add up how many ways our opponent can get all the hands that we just calculated. This will determine how many ways the opponent can beat Shinobi in this round of poker.

Full house: 257 ways Four of a Kind: 232 ways Straight Flush: 27 ways Royal Flush: 3 ways 257 + 232 + 27 + 3 = 519

Shinobi has obtained his flush in the suit of hearts. His five cards are: three (3), five (5), nine (9), Jack, and Queen. Due to the constant bluffing of the two suspicious guests, the other players except for Shinobi have folded their hands. Shinobi absolutely does not want to give them even an ounce of his money. If Shinobi decides not to fold and keep his flush, how many ways can his opponent get a hand to beat his flush? There are 519 ways that Shinobi's opponent can get a hand to beat his flush.

Question 5 Part B Solution

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Question 5 Part B Solution