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Math Question
- Thread starter emt1
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order matter?
Btw the solution is really simple.
For the first one it is (assuming order does not matter)
5 nCr 12
2nd part.
5 nCr 8
If order does matter it is still really simple.
don't fully understand your question, but you could have:
up to 2 'all face' hands
and 1 'all face but no king' hands...
12 face cards per pack (3*4) = 12
12/5 = 2.4
round down to 2
8 face cards (without king) per pack ((3*4)-4) = 8
8/5 = 1.6
round down to 1
up to 2 'all face' hands
and 1 'all face but no king' hands...
12 face cards per pack (3*4) = 12
12/5 = 2.4
round down to 2
8 face cards (without king) per pack ((3*4)-4) = 8
8/5 = 1.6
round down to 1
so if you don't count kings or ace's, that means your are only using queens and jacks (8 cards). 4 queens, 4 jacks = 8 total cards.
(If you are counting suit and the order of the cards matters) then.
1st card - can be 8 possibilities
2nd card - 7 possibilities
3rd card - 6 possibilities
4th card - 5 possibilities
5th card - 4 possibilities.
8x7x6x5x4 = 6720 possibilites.
but you really don't need them in any particular order and the suites are totally irrelevant because they are all going to be Full-Houses or 4 of a kind with a kicker...
So, if your question is how many different poker hands can you have...well
qqqqj (4 queens, jack kicker)
qqqjj (full house, queens over jacks)
qqjjj (full house, jacks over queens)
qjjjj (4 jacks queen kicker)
realistic answer is (4)
(If you are counting suit and the order of the cards matters) then.
1st card - can be 8 possibilities
2nd card - 7 possibilities
3rd card - 6 possibilities
4th card - 5 possibilities
5th card - 4 possibilities.
8x7x6x5x4 = 6720 possibilites.
but you really don't need them in any particular order and the suites are totally irrelevant because they are all going to be Full-Houses or 4 of a kind with a kicker...
So, if your question is how many different poker hands can you have...well
qqqqj (4 queens, jack kicker)
qqqjj (full house, queens over jacks)
qqjjj (full house, jacks over queens)
qjjjj (4 jacks queen kicker)
realistic answer is (4)
the fact that its a 52 cards deck is irrelevant...you are just asking how many 5 card hands can you have with 8 cards.
Again my answer is the same. What I just told you was the solution to calculated it. I could punch that in on a calculator and have the complete answer in less than 5 min.
Now if you are not allowed to use the nCr funtion on the caclulator it would take me 5 min to look up the formula nCr and then still less than 5 min to do it.
The solution to your question took more time for you to type it out than it would of taken to do.
So quick question: is this asking the probability of getting a hand of all face cards or the total number of combinations that will satisfy the conditions?
If it's about the total probability of getting a five-hand card with five face cards, the calculation is pretty easy:
12/52*11/51*10/50*9/49*8/48=95040/311875200=33/108290
If it's about the total number of combinations, I believe the formula is 12!/5!. It's been a while since I've taken probability, but I believe that's how you can find the total number of combinations (someone please correct that if it's wrong).
For the probability of pulling five queens or jacks, you should use the same method except with 8 (so 8/52*7/51...) and the combo formula should be 8!/5! (again, assuming this is the correct formula).
the question was, "how many 5-card hands will have all face cards?" (but not a king)
Perhaps my wording was vague, but omitting kings is the same as retaining queens and jacks.
If you'll notice, the multiplication I've used incorporates both queens and jacks into the probability equation.
Wow, a few wrong answers up there.
If face cards are jacks, queen, kings, then the number of 5 card hards that are all face cards is
12 nCr 5 = 12!/5!7! = 792, since there are 12 face cards and you want 5 of them.
The probability of getting a 5 card hand made entirely of face cards is the above number divided by 52 nCr 5 since that is the total number of 5 card hands that can be made.
thus 12 nCr 5 / 52 nCr 5 = (12!/5!7!) / (52!/5!47!) = 12!47!/7!52! = 0.000305
If you don't want kings, only jacks and queens then replace the 12 nCr 5 with 8 nCr 5 = 56.
If face cards are jacks, queen, kings, then the number of 5 card hards that are all face cards is
12 nCr 5 = 12!/5!7! = 792, since there are 12 face cards and you want 5 of them.
The probability of getting a 5 card hand made entirely of face cards is the above number divided by 52 nCr 5 since that is the total number of 5 card hands that can be made.
thus 12 nCr 5 / 52 nCr 5 = (12!/5!7!) / (52!/5!47!) = 12!47!/7!52! = 0.000305
If you don't want kings, only jacks and queens then replace the 12 nCr 5 with 8 nCr 5 = 56.
I knew I was forgetting something with my factorials, but to be honest, I can't remember why 7! needs to be in there.
Would you mind explaining for my sake?
I knew I was forgetting something with my factorials, but to be honest, I can't remember why 7! needs to be in there.
I think the question should be, why is the 5! there.
If there are 12 cards, and you need to choose 5, and the order in which you choose them matters, then the number of ways of doing that is 12*11*10*9*8 = 12! / 7!
Now, if the order in which you select the 5 cards doesn't matter, then the number of ways is (12! / 7!) / 5! = 12! / 5!7! (since there are 5! ways of choosing the same 5 cards)
The first case is the number of permutations, the second is the number of combinations. Permutations - order matters. Combinations - order doesn't matter.
Ahh, it's coming back to me slowly now.
I'll just take credit for remembering that the "!" is used for factorials and be on my way now.
Thanks for the explanation.
considering we are using playing cards, the order shouldn't matter because no matter how you look at the hand its going to equal the same thing:
example: 4 queens with a jack kicker is always going to read as "4 queens with a jack kicker" no matter what combination you put it in.
granted you can have 4 different kickers (jacks) so i guess that could change, but its irrelevant.
this is a silly question.
example: 4 queens with a jack kicker is always going to read as "4 queens with a jack kicker" no matter what combination you put it in.
granted you can have 4 different kickers (jacks) so i guess that could change, but its irrelevant.
this is a silly question.
The order doesn't matter, I was simply explaining where the formula comes from. And there are 4 ways of getting 4 kings with a jack kicker.
ok...yeah, i wasn't saying there was anything wrong with anyones math...i was just looking at it from a card players point of view...