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### Poker Combinations

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Date: 05/09/97 at 08:04:55
From: David Fried
Subject: Re: High School Precal (probability)

Dr. Math,

In a standard deck of cards, how many different ways are there for
getting a straight, a flush, a straight flush, one pair, or two pairs?

A straight consists of five cards in a row: (4,5,6,7,8) (A,2,3,4,5),
(10,J,Q,K,A). Note: The "A" cannot be in the middle.

A flush consists of five cards of the same suit: 5 clubs, 5 hearts,
and so on.

A straight flush consists of five cards of the same suit, in a row:
(4,5,6,7,8), all hearts, for example.

Two pairs consists of two separate pairs of two cards: 2 K's and 2 A's
or 2 4's and 2 10's

One pair consists of one pair of two cards.

Could you give the proofs please?

Thank you!
```

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Date: 05/14/97 at 08:09:06
From: Doctor Anthony
Subject: Re: High School Precal (probability)

How many different ways can you get:

(1) A straight - i.e., 4, 5, 6, 7, 8, (mixed suits).

The number of consecutive groups of 5 cards is 10. i.e., We can start
with A, 2, 3, ....., 10.

In each sequence of 5 we can choose each card in 4 ways (Clubs,
Hearts,...), this gives  4^5 ways = 1024.

So the total is 1024 x 10 = 10240. However, this total will include
the straight flush, so we must subtract the number that are straight
flushes. For the straight flush, the suit can be chosen in 4 ways.
So the number of straight flushes is 10 x 4 = 40. The number of
straights is 10240 - 40 = 10200.

(2) Flush (i.e. 5 cards of same suit).

You can choose the suit in 4 ways, and you can choose 5 cards from 13
in 13_C_5 = 1287 ways.

The total number of ways of getting a flush is 4 x 1287 = 5148. Again
we must subtract from this the number of ways of getting a straight
flush.

The number of flushes is 5148 - 40 = 5108.

(3) Straight flush.   As we have seen, there are 40.

(4) One pair (i.e., 2 K's, or 2 6's  etc.).

Suppose we have 2 K's.  The K's can be chosen in 4_C_2 ways = 6 ways

The face value of the pair can be chosen in 13 ways. Now we need to
choose 3 more cards, none of which match each other. So we choose 3 cards
out of the 12 remaining values, which is 12_C_3 = 220 ways. But also each
card can be 1 of the 4 suits, and so there are 4*4*4 ways to choose the suits.

Total number of ways is 6 x 13 x 220 x 4 x 4 x 4 = 1,098,240.

(5) Two pairs (i.e., 2 J's and 2 8's).

We can choose the 2 J's in 4_C_2 ways = 6 ways
We can choose the 2 8's in 4_C_2 ways = 6 ways

We can choose the face value of the pairs in 13_C_2 = 78 ways
We can choose the face value of fifth card in 11 ways, in each of the
four suits, for 44 ways.

Total number of ways is 6 x 6 x 78 x 11 x 4 = 123,552

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
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Associated Topics:
High School Permutations and Combinations

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