# Math Help - Help with deck of cards probability

1. ## 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)

2. 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)

poker two pair probability

You will get everything you want and more.

3. Hello, cb22hawk!

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.

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

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