Probability Question!?

A five-card poker hand is dealt and the last card is turned face up. Determine the probability that you have been dealt four aces, given that the card turned over is an ace. (A: 1/20825)

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  • 7 years ago
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    Okay, so we've got an ace turned up and four cards turned down, and we want to know the probability that all three of the other aces are in the four down cards.

    There are 51 possible cards (all but the ace that's showing) that could be in that 4-card stack. So there are

    C(51,4) = 51! / (47! 4!)

    = 51 * 50 * 49 * 48 / (4 * 3 * 2 * 1)

    = 51 * 50 * 49 * 2

    = 249,900 possible combinations in those 4 cards.

    [If you need any help with basic combinations formulas, I recommend the site linked from the source section below as a reference.]

    Now we need to know how many of the combinations in the 4 down cards could include the other three aces. Each of these must be a combination of the aces and one card that's not an ace, and there are 48 cards in the deck aside from the aces. So there are 48 combinations that give us the four-aces hand, and the probability is

    48/249,900 = 4/20,825

    = about 0.000192

    = roughly 0.02% (1/50 of 1%)

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