Question 1142759: 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 18.
ii) One of 5 different suits.
Hence, there are 90 cards in the deck with 18 ranks for each of the 5 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
43949268
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 16, 17, 18, 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 16, 17, 18, 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 by Alan3354(69443)   (Show Source):
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