Suited Blackjacks

Discussion in 'Sidewalk Cafe' started by mociferous, Mar 1, 2006.

  1. S. Yama

    S. Yama Active Member

    great formula

    Tirle_bj and Arlalik,

    P = 8N / 13(52N-1) - this is really very elegant formula.
    I don’t recall seeing it anywhere. Bravo!
    Do you care to share your train of thoughts on how you came up with it?

    I was familiar with Peter Griffin’s formula for blackjack (and suited would be a forth of it) for any subset of cards left:

    a * t * (1-2(a-1)*(t-1)/(n-2)/(n-3))/n/(n-1)

    where:
    a = number of aces,
    t = number of tens,
    n = cards remaining.
     
  2. Monkeysystem

    Monkeysystem Top Member Staff Member

    This Kept Me Up All Night

    Prob. of getting an Ace = 1/13
    Prob. of getting suit Ten to that Ace = 4N/(52N-1)

    4N is the number of suited Aces. (52N-1) is the number of remaining cards after removing the first Ace.

    You multiply the two probabilities together.
    Therefore Prob. of A-T suited starting with the Ace = 4N/13(52N-1)

    Prob. of getting a Ten = 4/13
    Prob. of getting an Ace suited to that Ten = N/(52N-1)
    You multiply the two probabilities together.
    Therefore Prob. of T-A suited starting with the Ten = 4N/13(52N-1)

    It's a combination, so you have to add the two probabilities together.

    So 4N/13(52N-1) + 4N/13(52N-1) = 8N/13(52N-1)

    Nice work, guys.
     
  3. mociferous

    mociferous Member

    Algebra in the real world

    Thanks, Monkeysystem et al... This gives me proof for my high school kids that algebra is helpful in the real world!
     
  4. tirle_bj

    tirle_bj Member

    thank you

    Thank you all, guys. Good analysis, Monkeysystem!
    Yama, here is our way to that formula:
    For Probabilities in Denominator we should have all possible combinations of two cards:
    [52N * (52N-1)] / 2 = 26N * (52N-1).
    In Numerator we should have total number of combinations with suited BJs.
    We have 4 suits, N Aces in each suite, and 4N Ten-value cards in each suite. So total we have: 4 * N * 4N = 16(N^2)
    The final probability is [16(N^2)] / [26N * (52N - 1)]
    To simplify we just divide numerator and denominator by 2N and get...
    8N / 13*(52N -1)
    Now for Odds we should just turn this formula up side down and here we go:
    13*(52N-1) / 8N.
    Again it's always pleasure to share opinions and thoughts between each other here on this wonderful site.
     
    Last edited: Mar 3, 2006

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