Total Free Money Earned

Redeems: $280,439

BTC Rate: $95945.67

Results 1 to 10 of 10
  1. #1
    Seasoned Veteran
    Join Date
    Dec 2012
    Posts
    51

    A poker formula question

    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.

  2. #2
    PokerOwned God abwil2's Avatar
    Join Date
    Mar 2011
    Posts
    2,672
    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

  3. #3
    Seasoned Veteran
    Join Date
    Feb 2013
    Posts
    37
    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%

  4. #4
    PokerOwned Demi-God HopsBar28's Avatar
    Join Date
    May 2013
    Posts
    1,031
    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

  5. #5
    PokerOwned Pro
    Join Date
    Sep 2013
    Posts
    197
    Lol yeah basically the guy with 109 is fkd. LOl

  6. #6
    Elite PokerOwned Member
    Join Date
    Oct 2013
    Posts
    797
    it is less than 1 percent of him catching the nine

  7. #7
    PokerOwned God
    Join Date
    Jan 2011
    Posts
    1,596
    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 (52-4)
    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)

  8. #8
    Library Master Champion eqgh5uea's Avatar
    Join Date
    Apr 2011
    Posts
    2,534
    Quote Originally Posted by HopsnBarley28 View Post
    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
    Fuck, can you send me a link to get that info?
    "We have met the enemy and they are ours; two ships, two brigs, one schooner and one sloop." --- O.H. Perry

  9. #9
    Library Master Champion eqgh5uea's Avatar
    Join Date
    Apr 2011
    Posts
    2,534
    Quote Originally Posted by bmeelneg View Post
    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 (52-4)
    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)
    That explains it.
    "We have met the enemy and they are ours; two ships, two brigs, one schooner and one sloop." --- O.H. Perry

  10. #10
    Seasoned Veteran
    Join Date
    Oct 2013
    Posts
    34
    Never really was good at math

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •