I'm curious how to figure out the odds of a flop coming, say, 10 9 9, when one person holds 1010 and the other 109.
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Thread: A poker formula question

11202013, 12:16 AM #1
A poker formula question

11202013, 08:01 AM #2
I have no clue but im sure someone here will pop ion and give you one!
Failing to Prepare is Preparing to fail : John Wooden

11202013, 04:31 PM #3
 Join Date
 Feb 2013
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if those are the only cards we are aware of, wouldn't it be (4/52)(3/51)(2/50)(4/52)(3/51)(2/50) = 0.00000328%

11202013, 06:40 PM #4
No.. there's only 1 10 left and 3 nines left in 48 cards.
So it's (1/48) x (3/47) x (2/46) x 100% = 0.0058 % chance. About 1 in 17000 chance of that happening

11202013, 06:47 PM #5
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Lol yeah basically the guy with 109 is fkd. LOl

11202013, 09:17 PM #6
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it is less than 1 percent of him catching the nine

11212013, 06:06 PM #7
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 Jan 2011
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here is how you figure this out ...
52 cards in a deck
4 cards known (ten ten , ten nine)
48 cards left in the deck preflop (524)
4 "good" cards (1 ten, 3 nine)
first card of the flop has a 4 out of 48 chance
now there is one less "good" card and one less card in the deck
second card has a 3 out of 47 chance
now there is one less "good" card and one less card in the deck
third card has a 2 out of 46 chance

(4/48) * (3/47) * (2/46)

11222013, 10:59 AM #8

11222013, 11:01 AM #9

11222013, 11:46 AM #10
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 Oct 2013
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Never really was good at math