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Manager
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poker q [#permalink] New post 30 Aug 2003, 11:19
hi all... i got this question from another forum of this site...
in poker, what's the prob to get 3 cards of a kind in a 5 cards hand?

my reasoning

1. we pick a card out of 13 types: 13c1=13
2. we pick 2 cards of the same type: 3c2=6
3. we pick 2 cards of different types: 12c1*12c1=12*12

so prob=13*6*12*12/(52c5)

comments? cheers, javier
Manager
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 [#permalink] New post 30 Aug 2003, 11:56
still thinking... id say i was wrong

1. we pick a card out of 13 types: 13c1=13
2. we pick 2 cards of the same type: 3c2=6
3. we pick 2 cards of different types: 48c2=48*47/2

so prob=(13*6*48*47/2)/(52c5)?
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 [#permalink] New post 30 Aug 2003, 15:08
alas, ive been revising my job and would say i was wrong... again! aaargh... well, heres my last try:
1. pick 1 card: 52; 2. find two that match: 52*3*2; since order doesnt matter: 52*3*2/3!
2. pick 2 other cards: 48*47/2!
answer: (52*3*2/3!)*(48*47/2!)/(52c5)
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 [#permalink] New post 03 Sep 2003, 11:53
Do you have the actual choices to the question?
Senior Manager
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 [#permalink] New post 04 Sep 2003, 06:47
Javier: I agree with your last post.

My approach was: [ 13*C(4,3)*C(48,2) ] / C(52,5) = 58656/2598960=2.25% aprox.
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 [#permalink] New post 04 Sep 2003, 09:04
Got it now with Martin's breakout :)
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 [#permalink] New post 04 Sep 2003, 10:23
nm. had a question about another problem listed in another topic. i guess i didnt read it correctly. sorry.
  [#permalink] 04 Sep 2003, 10:23
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