Question 1070086
<pre>
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
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a) How many different ways are there to get any 5 card hand<pre>

96 cards choose 5 = 96C5
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b)How many different ways are there to get exactly 1 pair (i.e. 2 cards with the same rank)?
<pre>
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
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c) How many different ways are there to get exactly 2 pair (i.e. 2 different sets of 2 cards with the same rank)?
<pre>
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
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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)?
<pre>
"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
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e) How many different ways are there to get exactly 4 of a kind (i.e. 4 cards with the same rank)?
<pre>
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
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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)?
<pre>
12 ranks choose 1 for the 5 of a kind = 12C1
8 suits choose 5 for the 5 cards = 8C5

Answer: 12C1*8C5
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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)?
<pre>
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
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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)?
<pre>
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
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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.
<pre>
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 
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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.
<pre>
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
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Remember: to find any of the probabilities just divide the answers by 96C5

Edwin</pre>