Anonymous
Anonymous asked in Science & MathematicsMathematics · 3 years ago

What is the probability that a five-card poker hand contains the ace of diamonds and the ace of hearts?

Update:

I am stuck in this question. Any help would be appreciated.

3 Answers

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  • Pope
    Lv 7
    3 years ago
    Favorite Answer

    number of possible hands = C(52, 5) = 52! / (47! 5!)

    number of possible successful hands = C(50, 3) = 50! / (47! 3!)

    probability of success

    = C(50, 3) / C(52, 5)

    = [50! / (47! 3!)] / [52! / (47! 5!)]

    = (50! 5!) / (52! 3!)

    = (5)(4) / [(52)(51)]

    = 5 / [(13)(51)]

    = 5 / 663

    ≈ 0.00754

  • 1 * 1 * 50 * 49 * 48 = number of hands with Ace of Diamond and the Ace of Hearts

    52! / (5! * 47!) = number of possible hands

    (50! / 47!) / (52! / (5! * 47!)) =>

    50! * 5! * 47! / (47! * 52!) =>

    50! * 5! / 52! =>

    50! * 5! / (50! * 51 * 52) =>

    5! / (51 * 52) =>

    5 * 4 * 3 * 2 / (51 * 52) =>

    5 * 3 * 2 / (51 * 13) =>

    5 * 2 / (17 * 13) =>

    10 / 221

    • Luiseezy3 years agoReport

      2 different answers. I don't know which of you is right. :(

  • Anonymous
    3 years ago

    I think I'm figuring this right. You have five chances out of 52 for the first ace and 4 chances out of 51 for the second ace. 5/52 x 4/51= 0.753%, so about three-quarters of one percent.

    • Lv 7
      3 years agoReport

      Damn, I was right and you chose the Johnny come lately. Thanks a lot.

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