Question 1070086: 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 12.
ii) One of 8 different suits.
Hence, there are 96 cards in the deck with 12 ranks for each of the 8 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
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
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
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
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
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
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
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 10, 11, 12, 1, 2)?
The number of ways of getting a straight flush is
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
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 10, 11, 12, 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
What is the probability of being dealt a straight that is not a straight flush?
Round your answer to 7 decimal places.
Answer by Edwin McCravy(20055)   (Show Source):
You can put this solution on YOUR website!
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
96 cards choose 5 or 96C5
-----------------------------------------------------------------
a) How many different ways are there to get any 5 card hand
96 cards choose 5 = 96C5
---------------------------------------------------------------
b)How many different ways are there to get exactly 1 pair (i.e. 2 cards with the same rank)?
12 ranks choose 1 for pair = 12C1
11 other ranks choose 3 = 11C3
8 suits choose 1 for the lowest ranking other card
8 suits choose 1 for the middle ranking other card
8 suits choose 1 for the highest ranking other card
12C1*11C3*8C1*8C1*8C1
-------------------------------------------------------------
c) How many different ways are there to get exactly 2 pair (i.e. 2 different sets of 2 cards with the same rank)?
12 ranks choose 2 for the two pairs = 12C2
8 suits choose 2 for the lower ranking pair = 8C2
8 suits choose 2 for the higher ranking pair = 8C2
Answer: 12C2*8C2*8C2
-----------------------------------------------------------
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.
12 ranks choose 1 for the 3 of a kind = 12C1
8 suits choose 3 for the suits of the 3 of a kind = 8C3
11 other ranks choose 2 for the other two = 11C2
8 suits choose 1 for the lower ranking other card = 8C1
8 suits choose 1 for the higher ranking other card = 8C1
Answer: 12C1*8C3*11C2*8C1*8C1
--------------------------------------------------------
e) How many different ways are there to get exactly 4 of a kind (i.e. 4 cards with the same rank)?
12 ranks choose 1 for the 4 of a kind = 12C1
11 other ranks choose 1 for the 5th card = 11C1
8 suits choose 1 for the 5th card = 8C1
Answer: 12C1*11C1*8C1
------------------------------------------------------
f) How many different ways are there to get exactly 5 of a kind (i.e. 5 cards with the same rank)?
12 ranks choose 1 for the 5 of a kind = 12C1
8 suits choose 5 for the 5 cards = 8C5
Answer: 12C1*8C5
----------------------------------------------------
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)?
12 ranks choose 1 for the 3 of a kind = 12C1
8 suits choose 3 for the 3 of a kind = 8C3
11 other ranks choose 1 for the pair = 12C1
8 suits choose 2 for the pair = 8C2
Answer: 12C1*8C3*11C1*8C2
-------------------------------------------------
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, 11, 12, 1, 2)?
The straight flush can have any of these 8 sequences of ranks
1 thru 5, 2 thru 6, 3 thru 7, 4 thru 8, 5 thru 9, 6 thru 10, 7 thru 11, or 8 thru 12
That's
8 sequences of ranks choose 1 = 8C1
8 suits choose 1 = 8C1
Answer: 8C1*8C1
------------------------------------------------
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.
8 suits choose 1 (for all 5 cards) = 8C1
12 ranks choose 5 = 12C5
8C1*12C5
Subtract 8C1*8C1 straight flushes
Answer: 8C1*12C5 - 8C1*8C1
--------------------------------------------
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, 11, 12, 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 8 sequences of ranks
1 thru 5, 2 thru 6, 3 thru 7, 4 thru 8, 5 thru 9, 6 thru 10, 7 thru 11, or 8
thru 12
That's
8 sequences of ranks choose 1 = 8C1
8 suits choose 1 for the lowest ranking card = 8C1
8 suits choose 1 for the next to lowest ranking card = 8C1
8 suits choose 1 for the middle ranking card = 8C1
8 suits choose 1 for the next to highest ranking card = 8C1
8 suits choose 1 for the highest ranking card = 8C1
Subtract 8C1*8C1 straight flushes
Answer 8C1*8C1*8C1*8C1*8C1 - 8C1*8C1
------------------------------------------
Remember: to find any of the probabilities just divide the answers by 96C5
Edwin
|
|
|
