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TeHCM
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TeHCM
Sorry - Full house = 3 cards of the same face value + 2 cards of the same value :oops:


Geez....I'm not good at card games :oops:
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Full house can be = 3 same face value + any 2 cards of same value

Consider case of holding 3 cards with face value 1, then the other 2 cards cannot be of face value 1.

So (1,1,1) and (2,2) or (3,3) or (4,4)... (A,A)
(1,1,1) -> 4C3 = 4 ways
(2,2) -> 4C2 = 6 ways
(2,2).. (A,A) = 6 * 13 = 78

Total number of ways to pick Full house One = 4 * 78 = 312
We can have full house 1,2,3,4,5,6,7,8,9,10,J,Q,K,A - this is 4368 ways

Number of ways to pick 5 cards = 52C5 = 2598960

P = 4368/2598960 = 1092/649740 = 273/162435 = 1/595
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gmatbangkok
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Is there a card value of 1??

Anyway, I am trying to rewrite the formula again.

Prob = (14C1 * 13C1 *4C3 *4C2)/ (52C5)
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kfranson
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We have 13 denominations (2 thru Ace) and we need to choose two of them to make the full house, so that's 13c2.

We will choose 3 from one of the denominations, 4c3 and 2 from the other 4c2.

Keep in mind that you can have 3 sixes and 2 sevens or 2 sixes and 3 sevens so in the 13c2 above we can actually remove the 2 from the denominator or simply multiply by 2 at the end (in other words 67 is different from 76, order matters b/c we could have 3 sixes and 2 sevens or vice versa).

So ways to make a full house = (13c2)*(4c3)*(4c2)*2.
Now divide this by 52*51*50*49*48 to get the probability.



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