# Help with deck of cards probability

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• Sep 4th 2008, 10:56 PM
cb22hawk
Help with deck of cards probability
I have been working on these problems for a several hours and am stuck. Can anyone help me at all? Here are the problems:

A poker hand is defined as drawing five cards at random without replacement from a deck of 52 playing cards. Find the probability of each of the following poker hands:

A. Two pairs (two pairs of equal face value plus one card of a different value)
B. One pair (one pair of equal face value plus three cards of different values)
• Sep 5th 2008, 05:30 AM
mr fantastic
Quote:

Originally Posted by cb22hawk
I have been working on these problems for a several hours and am stuck. Can anyone help me at all? Here are the problems:

A poker hand is defined as drawing five cards at random without replacement from a deck of 52 playing cards. Find the probability of each of the following poker hands:

A. Two pairs (two pairs of equal face value plus one card of a different value)
B. One pair (one pair of equal face value plus three cards of different values)

Google:

poker two pair probability

You will get everything you want and more.
• Sep 5th 2008, 07:25 AM
Soroban
Hello, cb22hawk!

Quote:

A poker hand is defined as drawing five cards at random
without replacement from a deck of 52 playing cards.

There are: . ${52\choose5} \:=\:2,598,960$ possible poker hands.

Quote:

Find the probability of each of the following poker hands:

A. Two pairs (two pairs of equal face value plus one card of a different value)

There are: . ${13\choose2} = 78$ choices for the values of the pairs.

There are: . ${4\choose2}{4\choose2} \:=\:6^2$ ways to get the two pairs.

And there are $44$ choices for the fifth card.

Hence, there are: . $78 \times 6^2 \times 44 \:=\:123,552$ ways to get two pairs.

Therefore: . $P(\text{two pairs}) \:=\:\frac{123,552}{2,598,960} \:=\:\frac{198}{4165}$

Quote:

B. One pair (one pair of equal face value plus three cards of different values)

There are: . $13$ choices for the value of the pair.

There are: . ${4\choose2} = 6$ ways to get the pair.

There are: . $\frac{48\cdot44\cdot40}{3!} = 14,080$ choices for the other three cards.
. .
(They must not contain another pair.)

Hence, there are: . $13 \times 6 \times 14,080 \:=\:1,098,240$ ways to get one pair.

Therefore: . $P(\text{one pair}) \;=\;\frac{1.098,240}{2.598,960} \;=\;\frac{352}{833}$