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# poker q

Author Message
Manager
Joined: 14 Aug 2003
Posts: 88
Location: barcelona

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30 Aug 2003, 12:19
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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)

Manager
Joined: 14 Aug 2003
Posts: 88
Location: barcelona

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30 Aug 2003, 12:56
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)?
Manager
Joined: 14 Aug 2003
Posts: 88
Location: barcelona

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30 Aug 2003, 16:08
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!
Manager
Joined: 20 Apr 2003
Posts: 53

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03 Sep 2003, 12:53
Do you have the actual choices to the question?
Senior Manager
Joined: 22 May 2003
Posts: 329
Location: Uruguay

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04 Sep 2003, 07:47
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.
Manager
Joined: 20 Apr 2003
Posts: 53

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04 Sep 2003, 10:04
Got it now with Martin's breakout
Manager
Joined: 20 Apr 2003
Posts: 53

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04 Sep 2003, 11:23
nm. had a question about another problem listed in another topic. i guess i didnt read it correctly. sorry.
04 Sep 2003, 11:23
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