SOLUTION: 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 b

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Question 1187855: 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 16.
ii) One of 9 different suits.
Hence, there are 144 cards in the deck with 16 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.
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.

Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
b)How many different ways are there to get exactly 1 pair (i.e. 2 cards with the
same rank)?

To get one pair, we first:

Choose the 1 rank.  That's 16 ranks choose 1, 16C1=16 ways.

Then we choose the 2 suits.  That's 9 suits choose 2, 9C2=36 ways.

We choose the ranks of the other three cards different from the 1 rank that the
pair has, and also different from each other, so there will not be another pair:
That's 15 other ranks choose 3, 15C3 = 455 ways.

We choose the suit for the card of the other 3 with the lowest rank 9 ways.
We choose the suit for the card of the other 3 with the middle rank 9 ways.
We choose the suit for the card of the other 3 with the highest rank 9 ways.

That's (16C1)(9C2)(9)(9)(9) = (16)(36)(9)(9)(9) = 419904

So, the number of ways of getting exactly 1 pair is 419904 ways. 

The number of ways to get any 5 cards is 144C5 = 481008528

So the probability is 419904 out of 481008528 or 419904/481008528 which
reduces to 2916/3340337 or 0.0008730, to 7 decimal places.

Edwin
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