1. Welcome to VegasMessageBoard
    It appears you are visiting our community as a guest.
    In order to view full-size images, participate in discussions, vote in polls, etc, you will need to Log in or Register.

Video Poker Need some math wizards help

Discussion in 'Video Poker' started by FLACookieman, Sep 29, 2016.

  1. FLACookieman

    FLACookieman Tourist

    Joined:
    May 9, 2010
    Messages:
    10
    Location:
    Jacksonville
    Trips to Las Vegas:
    35
    I just came back from Vegas. I was playing Deuces Wild Triple Play. Was dealt 3 deuces on bottom hand, of course i held the deuces and hit the draw button. To my amazement all 3 hands filled in the 4th deuce !!!!
    I would like to know the odds of this happening or what the probability is of this happening.
    Thanks in advance.
     
  2. Wedgefromhere

    Wedgefromhere Tourist

    Joined:
    Jan 9, 2010
    Messages:
    88
    Trips to Las Vegas:
    25
    If I did this correctly the probably of getting a deuce in all three rows given you already got three deuces is 0.0077%.
     
  3. vetsen

    vetsen Tourist

    Joined:
    Sep 26, 2009
    Messages:
    43
    Trips to Las Vegas:
    23
    that's correct, about 1 in 13000.
     
  4. tringlomane

    tringlomane STP Addicted Beer Snob

    Joined:
    Jan 21, 2013
    Messages:
    9,890
    Location:
    Missouri
    Trips to Las Vegas:
    15
    Yes, as others have said, converting any dealt trips to quads on all 3 lines is:

    (2/47)^3 ~ 0.000077 ~ 1 in 12,978

    Nice pull!

    And I'm glad you seem a lot more excited than the only time I saw it (via a neighbor). He pulled quad Jacks on DDB on each of the 3 lines, doesn't react at all, and immediately moves onto the next hand. He really sucked the joy out of that moment.
     
    • Like Like x 1
  5. TXactuarial

    TXactuarial Low-Roller

    Joined:
    Feb 6, 2014
    Messages:
    324
    Location:
    Iowa
    Trips to Las Vegas:
    6
    Probability is simultaneously the cause of excitement and a buzzkill.
     
  6. Aces and Eights

    Aces and Eights VIP Whale

    Joined:
    Jul 24, 2014
    Messages:
    1,121
    Location:
    Southern California
    Trips to Las Vegas:
    100
    Correct me if I'm wrong, but isn't it (1/47+1/46)^3 ~ 0.000080? Wouldn't 2/47 be the odds of getting one of two specified cards drawing one card, where (1/47+1/46) is the odds of getting a specific card drawing twice? And of course to the third power for three different hands.
     
  7. Wedgefromhere

    Wedgefromhere Tourist

    Joined:
    Jan 9, 2010
    Messages:
    88
    Trips to Las Vegas:
    25
    If you save three cards (the three deuces), there are two cards to be drawn. To get four deuces, either the first of two cards has to be a deuce or the second has to be deuce. So the probability is (1/47) x (46/46) + (46/47) x (1/46) = 2/47. Since the three lines are independent, the final answer is (2/47)^3 = 0.000077.
     
  8. tringlomane

    tringlomane STP Addicted Beer Snob

    Joined:
    Jan 21, 2013
    Messages:
    9,890
    Location:
    Missouri
    Trips to Las Vegas:
    15
    A lot of times calculations like this do what you describe with card removal, but when you are looking at it the way you're describing, you are double counting during the times you hit the quad on the first card.

    Probability of getting the quad on the first card: 1/47
    Probability of getting the quad on the second card if you didn't get the quad on first card: 1/46
    Probability of getting the quad on the second card if you got the quad on the first card: 0 since there are only 4 cards of that rank in the deck.

    So using your method, the probability of getting a quad drawing two cards is like this:

    Probability of getting quad on 1st card + Probability of not getting quad on 1st card x Probability of getting quad on 2nd card

    1/47 + (46/47) x (1/46) = 1/47 + 1/47 = 2/47

    Alternatively you can find the probability this way...many multi card draws are calculated this way. Find the probability of it failing and subtract from 1.

    Probability of Quad = 1 - (Probability of not getting quad on 1st card) x (Probability of not getting quad on the 2nd card) = 1 - (46/47) x (45/46) = 1 - 45/47 = 2/47
     
  9. Aces and Eights

    Aces and Eights VIP Whale

    Joined:
    Jul 24, 2014
    Messages:
    1,121
    Location:
    Southern California
    Trips to Las Vegas:
    100
    Ah, got it! Thanx! It makes sense to me now because if you draw 47 times, the chances of getting the 4OAK should be 1 = 1/47 + 1/47 +1/47 + ... (47 times).

    So the odds of getting a 4OAK from a 3OAK is the same as getting a straight flush from an outside straight flush draw?
     
    Last edited: Sep 29, 2016
  10. tringlomane

    tringlomane STP Addicted Beer Snob

    Joined:
    Jan 21, 2013
    Messages:
    9,890
    Location:
    Missouri
    Trips to Las Vegas:
    15
    Yes. Both are 2 in 47 or 1 in 23.5.