Question

1) We are creating a new card game with a new deck. Unlike the normal deck that has 13 ranks (Ace through King) and 4 Suits (hearts, diamonds, spades, and clubs), our deck will be made up of the following.

Each card will have:
i) One rank from 1 to 10.
ii) One of 9 different suits.

Hence, there are 90 cards in the deck with 10 ranks for each of the 9 different suits, and none of the cards will be face cards! So, a card rank 11 would just have an 11 on it. Hence, there is no discussion of "royal" anything since there won't be any cards that are "royalty" like King or Queen, and no face cards!

The game is played by dealing each player 5 cards from the deck. Our goal is to determine which hands would beat other hands using probability. Obviously the hands that are harder to get (i.e. are more rare) should beat hands that are easier to get.a) How many different ways are there to get any 5 card hand?
The number of ways of getting any 5 card hand is  
DO NOT USE ANY COMMAS

b)How many different ways are there to get exactly 1 pair (i.e. 2 cards with the same rank)?
The number of ways of getting exactly 1 pair is
DO NOT USE ANY COMMAS

What is the probability of being dealt exactly 1 pair?
Round your answer to 7 decimal places.


c) How many different ways are there to get exactly 2 pair (i.e. 2 different sets of 2 cards with the same rank)?
The number of ways of getting exactly 2 pair is
DO NOT USE ANY COMMAS

What is the probability of being dealt exactly 2 pair?
Round your answer to 7 decimal places.


d) How many different ways are there to get exactly 3 of a kind (i.e. 3 cards with the same rank)?
The number of ways of getting exactly 3 of a kind is
DO NOT USE ANY COMMAS

What is the probability of being dealt exactly 3 of a kind?
Round your answer to 7 decimal places.


e) How many different ways are there to get exactly 4 of a kind (i.e. 4 cards with the same rank)?
The number of ways of getting exactly 4 of a kind is
DO NOT USE ANY COMMAS

What is the probability of being dealt exactly 4 of a kind?
Round your answer to 7 decimal places.


f) How many different ways are there to get exactly 5 of a kind (i.e. 5 cards with the same rank)?
The number of ways of getting exactly 5 of a kind is  
DO NOT USE ANY COMMAS

What is the probability of being dealt exactly 5 of a kind?
Round your answer to 7 decimal places.


g) How many different ways are there to get a full house (i.e. 3 of a kind and a pair, but not all 5 cards the same rank)?
The number of ways of getting a full house is  
DO NOT USE ANY COMMAS

What is the probability of being dealt a full house?
Round your answer to 7 decimal places.


h) How many different ways are there to get a straight flush (cards go in consecutive order like 4, 5, 6, 7, 8 and all have the same suit. Also, we are assuming there is no wrapping, so you cannot have the ranks be 8, 9, 10, 1, 2)?
The number of ways of getting a straight flush is  
DO NOT USE ANY COMMAS

What is the probability of being dealt a straight flush?
Round your answer to 7 decimal places.


i) How many different ways are there to get a flush (all cards have the same suit, but they don't form a straight)?
Hint: Find all flush hands and then just subtract the number of straight flushes from your calculation above.
The number of ways of getting a flush that is not a straight flush is
DO NOT USE ANY COMMAS

What is the probability of being dealt a flush that is not a straight flush?
Round your answer to 7 decimal places.


j) How many different ways are there to get a straight that is not a straight flush (again, a straight flush has cards that go in consecutive order like 4, 5, 6, 7, 8 and all have the same suit. Also, we are assuming there is no wrapping, so you cannot have the ranks be 8, 9, 10, 1, 2)?
Hint: Find all possible straights and then just subtract the number of straight flushes from your calculation above.
The number of ways of getting a straight that is not a straight flush is  
DO NOT USE ANY COMMAS

What is the probability of being dealt a straight that is not a straight flush?
Round your answer to 7 decimal places.

Answer 1

In these problems nCr means the number of combinations of n things
taken r at a time.

Also to get all the probabilities, just divide the number of successful hands by
the number of possible hands, which is
90 cards choose 5 or 90C5

a) How many different ways are there to get any 5 card hand

90 cards choose 5 = 90C543949268
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b)How many different ways are there to get exactly 1 pair (i.e. 2 cards with the same rank)?

10 ranks choose 1 for pair = 10C1
9 other ranks choose 3 = 9C3
9 suits choose 1 for the lowest ranking other card
9 suits choose 1 for the middle ranking other card
9 suits choose 1 for the highest ranking other card

10C1*9C3*9C1*9C1*9C1 = 612360
-------------------------------------------------------------

c) How many different ways are there to get exactly 2 pair (i.e. 2 different sets of 2 cards with the same rank)?

10 ranks choose 2 for the two pairs = 10C2
9 suits choose 2 for the lower ranking pair = 9C2
9 suits choose 2 for the higher ranking pair = 9C2

Answer:  10C2*9C2*9C2 = 58320
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d) How many different ways are there to get exactly 3 of a kind (i.e. 3 cards with the same rank)?

"Exactly" means the other two are not a pair and there are not 4 of the same rank. 

10 ranks choose 1 for the 3 of a kind = 10C1
9 suits choose 3 for the suits of the 3 of a kind = 9C3
9 other ranks choose 2 for the other two = 9C2
9 suits choose 1 for the lower ranking other card = 9C1
9 suits choose 1 for the higher ranking other card = 9C1

Answer:  10C1*9C3*9C2*9C1*9C1 = 2449440
-------------------------------- ------------------------

e) How many different ways are there to get exactly 4 of a kind (i.e. 4 cards with the same rank)?

10 ranks choose 1 for the 4 of a kind = 10C1
9 other ranks choose 1 for the 5th card = 9C1
9 suits choose 1 for the 5th card = 9C1

Answer: 10C1*9C1*9C1 = 8100
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f) How many different ways are there to get exactly 5 of a kind (i.e. 5 cards with the same rank)?

10 ranks choose 1 for the 5 of a kind = 10C1
9 suits choose 5 for the 5 cards = 9C5

Answer: 10C1*9C5 = 1260
----------------------------------------------------

g) How many different ways are there to get a full house (i.e. 3 of a kind and a pair, but not all 5 cards the same rank)?

10 ranks choose 1 for the 3 of a kind = 10C1
9 suits choose 3 for the 3 of a kind = 9C3
9 other ranks choose 1 for the pair = 9C1
9 suits choose 2 for the pair = 9C2

Answer: 10C1*9C3*9C1*9C2 = 272160
-------------------------------------------------

h) How many different ways are there to get a straight flush (cards go in consecutive order like 4, 5, 6, 7, 8
and all have the same suit. Also, we are assuming there is no wrapping, so you cannot have the ranks be 10, 1, 2)?

The straight flush can have any of these 6 sequences of ranks 

1 thru 5, 2 thru 6, 3 thru 7, 4 thru 8, 5 thru 9, 6 thru 10

That's
6 sequences of ranks choose 1 = 6C1
9 suits choose 1 = 9C1

Answer: 6C1*9C1 = 54
------------------------------------------------

i) How many different ways are there to get a flush (all cards have the same suit, but they don't form a
straight)?
Hint: Find all flush hands and then just subtract the number of straight flushes from your calculation above.

9 suits choose 1 (for all 5 cards) = 9C1
10 ranks choose 5 = 10C5

9C1*10C5

Subtract 6C1*9C1 straight flushes

Answer: 9C1*10C5 - 6C1*9C1 = 2214
--------------------------------------------

j) How many different ways are there to get a straight that is not a straight flush (again, a straight flush has cards that go in consecutive order like 4, 5,
6, 7, 8 and all have the same suit. Also, we are assuming there is no wrapping,
so you cannot have the ranks be 10, 1, 2)?
Hint: Find all possible straights and then just subtract the number of straight
flushes from your calculation above.

The straight can be any of these 6 sequences of ranks 

1 thru 5, 2 thru 6, 3 thru 7, 4 thru 8, 5 thru 9, 6 thru 10

That's
6 sequences of ranks choose 1 = 6C1
9 suits choose 1 for the lowest ranking card = 9C1
9 suits choose 1 for the next to lowest ranking card = 9C1
9 suits choose 1 for the middle ranking card = 9C1
9 suits choose 1 for the next to highest ranking card = 9C1
9 suits choose 1 for the highest ranking card = 9C1

Subtract 6C1*9C1 straight flushes

Answer 6C1*9C1*9C1*9C1*9C1 - 6C1*9C1 = 39312

to find any of the probabilities just divide the answers by 96C5