What is the probability of getting exactly one ace in a 5 card poker hand?

and why?

Thanks!

5 Answers

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  • Andy J
    Lv 7
    8 years ago
    Best Answer

    There are 53C5 = 2598960 possible 5-card hands.

    There are 4 aces and 48 non-aces. There are 4 ways to choose one ace, and 48C4 = 194580 ways to choose 4 more cards to fill out the hand from the non-aces. Total number of hands is 4 * 194580 = 778320.

    Probability is 778320 / 2598960 = 3243 / 10829 or about 0.2995.

    EDIT: x8oi, the question explicitly asks about *exactly* one ace, not *at least* one ace as you have calculated.

    EDIT 2: gjdolby, your answer would be correct if we cared about the order the cards were drawn. We don't, so we need to multiply your answer by the number of positions the ace could be drawn in, which is 5C1 = 5:

    0.059894727 * 5 = 0.299473635 agreeing with my answer.

  • Saul
    Lv 4
    4 years ago

    The chance of one ace being drawn is 4/52 on the first draw. If non drawn the next card is 4/51, then next 450 and 4/49 and 4/48.each time an ace is drawn the3 top number will reduce by one. So if the first card is an ace the next odds would be 3/51 and so on. To draw 2 aces in 5 cards is (5 x 2/4)/52 = 2.1/2 in 52. or just under 5%.

  • dotto
    Lv 4
    3 years ago

    Poker Hand Probability

  • 8 years ago

    1 card is one of four aces, and the other 4 cards are drawn from the remaining 48 non-aces.

    4/52 * 48/51 * 47/50 * 46/49 * 45/48 = 6486 / 108290 ≈ 0.059894727

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  • x8oi
    Lv 4
    8 years ago

    There are 52 non-joker cards in a pack.

    There are 4 aces in a pack.

    The probability of having 1 ace if dealt 1 card is 4/52 (~7.692%).

    For your second card (assuming the first card wasn't an ace), the probability is 4/51 (~7.842%)

    For your third card (assuming the first and second cards weren't an ace), the probability is 4/50 (8%)

    (etc)

    But... to work out the combined probabilities, you have to work out the probability that you *won't* be dealt an ace. Then you multiply them together. So:

    (48/52) * (47/51) * (46/50) * (45/49) * (44/48) = ~ 65.88%

    So the probability that you *will* be dealt an ace is 100% - 65.88%.

    Thus, the probability that at least one card in your hand is an ace is about 34%.

    EDIT: Andy J is right. I accept defeat. :-(

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