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what is the probability of getting full house (for example: [#permalink]
 14 Feb 2005, 19:32
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what is the probability of getting full house (for example: three kings and two jacks) from a deck of cards?
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GMAT Club Legend
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Not too sure what full house is. Does the 5 cards all have to be pictures ?? (as in, 3 queens, 2 jacks, or 3 jacks 2 queens)
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Total outcome: c(54,5)
Getting full house: C(13,2)*C(3,2)*C(3,2)
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HongHu wrote: Total outcome: c(54,5)
Getting full house: C(13,2)*C(3,2)*C(3,2)
try again.
Buddy goodluck for tomorrow's exam.
Last edited by MA on 19 Feb 2005, 23:47, edited 1 time in total.
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MA wrote: HongHu wrote: Total outcome: c(54,5)
Getting full house: C(13,2)*C(3,2)*C(3,2) try again. BODY goodluck for tomorrow's exam. I can't give the answer but I am wondering Hong Hu, why c(54,5) ? not c(52,5) for total outcome ?
For the full house we must consider the fact that if you have 3 kings first, you have to exclude the kings from the potential pair of king in the probability for the pair of the remaining 2 card because it would be impossible, 4 kings maximum in the game and not 5... 
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MA wrote: Antmavel wrote: HongHu wrote: I can't give the answer but I am wondering Hong Hu, why c(54,5) ? not c(52,5) for total outcome ? a valid question?  by re-reading myself I admit that my remark was not very clear, I write again now :
HongHu, in your post you've written c(54,5) as the formula to calculate the total outcome. why didn't you choose c(52,5) for this formula ?
Hope now it's clear for everybody 
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Manager
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MA wrote: what is the probability of getting full house (for example: three kings and two jacks) from a deck of cards?
now lets analyze...wat we want is five cards. So our total outcomes are:
52 52!
C = --------- = 2598960
5 47!*5!
Now what we want is possible outcomes for (1)3 similar and (2)2 similar cards.
(1)Firstly lets look at possibility of getting three similar cards. Definitely, three similar cards can come from four similar cards...not more than that. so the number of similar cards for a quartlets is:
4
C = 4 possibilities for getting 3 similar cards from a single type.
3
since there are 13 different types so we add 4 13 times or 4*13 = 52. Thus we have 52 possible ways of having three similar cards.
(2)Now for getting two similar cards.
4
C = 6 possiblities of two similar cards of a type.
2
And 6*12 = 72 for all 12 types, since one has been used in above 3 cards.
Now for the big equation which is:
52*72 3744 ~ 144
Probability = ------------ = ------------ = --------
2598960 2598960 10000
I dont think u wud eva get such a question on GMAT...since its calculation is hectic.
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MA wrote: Antmavel wrote: I can't give the answer but I am wondering Hong Hu, why c(54,5) ? not c(52,5) for total outcome ? a valid question? I was more used to counting the two jokers included.  C(52, 5) is good. 
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HongHu wrote: I was more used to counting the two jokers included. C(52, 5) is good. you r (still) missing the only and gr8 joker 
anyway,
prob of getting a full house =(13c3 x 12c2 x 4c3 x 4c2)/(52c2)
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MA wrote: HongHu wrote: I was more used to counting the two jokers included. C(52, 5) is good. you r (still) missing the only and gr8 joker anyway, prob of getting a full house =(13c3 x 12c2 x 4c3 x 4c2)/(52c2) Please spell out the numerator portion of your answer
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MA wrote: HongHu wrote: Total outcome: c(54,5)
Getting full house: C(13,2)*C(3,2)*C(3,2) try again. Buddy goodluck for tomorrow's exam. Hey I can't believe that I made a crucial typo.
It should be C(13,2)*C(3,2)*C(3,1).
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MA wrote: you r (still) missing the only and gr8 joker anyway, prob of getting a full house =(13c3 x 12c2 x 4c3 x 4c2)/(52c2) Ok let's compare numbers. I don't know why you'd do it your way. But my number would be: 13*6*3*3
Oh well that should be good enough for comparison. Don't want to multiply them out.
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I agree with Honghu's answer and I'll try to explain a bit further.
Total outcomes: 52C5 --> we have to select 5 out of 52 cards(excluding jokers)
Favorable outcomes:
step#1: Select 2 out of 13 sets (2,3,4,J,Q, etc...) --> 13C2
step#2: Select 2 out of set of 1st set of 4 cards given that 1st card was already picked. This will ensure that we pick a card different from the other set of 4 cards and this will be part of the group of 3 for the flush--> 3C2
step#3: Select 1 out of set of 2nd set of 4 cards given that 1st card was already picked. This is part of the pair in the flush --> 3C1
Answer is [13C2*3C2*3C1] / 52C5
_________________
Best Regards,
Paul
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Paul, I couldn't understand the concept of the question initially. But once I read your explanation, I got it. Let me tell you, this was an awesome explaination. You really make complicated things simple by writing them in simple english with emphasis on imp phrases.
It helps a lot.
Thanks
Vijo.
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