Question

From a deck of 52 cards, how many five card poker hands can be formed if there is a pair (two of the cards are the same number, and none of the other cards are the same number)?

The first part, selecting two cards of the same number is pretty simple,
\[{_4}C_2\]
I'm confused about the next bit though, as none of the other cards can be the same as each other and the pair. Supposedly I should subtract 4 cards first, 48, then another 4, etc, which makes sense, though I'm not sure what combinations I should use to represent the scenario. The final answer is 1098240.

Answer 1

I am not getting the right answer yet:

But a better way to think of the question would be:
you 4 cards with different numbers ... and 1 card with the same number as one of the other 4

Answer 2

ok ... got it!

Answer 3

u still here @finaiized ?

Answer 4

choose 4 cards with different numbers ... and 1 card with the same number as one of the other 4

Answer 5

Sorry @PaxPolaris, I'm back now!

Answer 6

first choosing 4 cards with different numbers:
\[\Large {52 \times 48 \times 44 \times 40 \over 4!}\]

Answer 7

do you follow so far @finaiized ?

Answer 8

Question: Why did you divide by (4!)? I just assumed choosing 4 different cards would be 52Cr1, 48Cr1, ...

Answer 9

we cannot use (n choose r) here ... because the numbers go down by 4

Answer 10

Okay, I see.

Answer 11

this is always true:
\[\text {number of combinations possible}={\text {number of permutations possible}\over \text {number of ways to order 1 possible combination}}\]

52*48*44*40 = number of permutations possible
4!= number of ways to order 1 possible combination

Answer 12

Oh yes, I remember that.

Answer 13

Next, the fifth card has 12 possibilities.

Answer 14

so, times 12.
do you follow @finaiized ?

Answer 15

Yep, 13 cards - the one that is already chosen, right?

Answer 16

the 5th card has to be one of the 4 numbers already chosen ....

there are 3 of each left ....

4*3=12

Answer 17

@finaiized ?

Answer 18

Oh... sorry, these cards are confusing me. Go on though, thank you for your continual explanations.

Answer 19

finally, let's look at an example:

the 1st 4 are:
2, 3, 4, 5-hearts.

the 5th is: 5-clubs.

Answer 20

this is the same as if the 1st 4 are:
2, 3, 4, 5-clubs.

the 5th is: 5-hearts.


So you need to divide by 2 to get the final answer.

Answer 21

52*48*44*40/4! *12/2 = 1098240

Answer 22

For the first 4, do you mean the one's chosen?

Answer 23

? yes... the 4 that cannot match

Answer 24

I'm still confused as to why you need to divide by two though. Why does choosing one of the same cards halve the total number of possibilities?

Answer 25

it took me a while to understand that too..

because, as i showed you in the example, there are 2 ways to get the same 5 cards....

Answer 26

gtg now ... bye

Answer 27

Oh, your examples were linked... ;)
Thanks, I believe I understand it now. Thank you for your time!

Answer 28

we could also have solved this without re-wording the question: (just wasn't thinking that way earlier)
how many five card poker hands can be formed if there is a pair (two of the cards are the same number, and none of the other cards are the same number)

you were almost on the right track :

you don't do \(\large _4C_2 =\Large{4\times3\over2!}\)
Instead, we have 52 choices for the 1st card, and 3 for the matching 2nd:
\[\Large = {52\cdot 3 \over 2!}\]

then multiply it by \[\Large \times {48\cdot44\cdot40 \over 3!}\] for the final 3 non-matching cards.