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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...

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.

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.

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

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.

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