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Math Help - Probability of getting a card hand dealt in a particular order.

  1. #1
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    Probability of getting a card hand dealt in a particular order.

    52 cards, dealt 5 cards
    P(1st 4 are aces, last is face card)
    the answer is 1/1082900
    how do i do it (besides the [4/52][3/51][2/50][1/49][12/48])?
    shouldn't 4 nCr 4 x 12 nCr 1 all over 52 nCr 5 work?????!?!?!
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by mairjuanaman View Post
    52 cards, dealt 5 cards
    P(1st 4 are aces, last is face card)
    the answer is 1/1082900
    how do i do it (besides the [4/52][3/51][2/50][1/49][12/48])?
    shouldn't 4 nCr 4 x 12 nCr 1 all over 52 nCr 5 work?????!?!?!
    That would work, if it were not for the fact that you are asked to determine the probability of being given those 5 cards in that particular sequence: 4 aces first, and only then one single face card. Also, if you are interested in the probability for a particular sequence of aces, you cannot use 4 nCr 4 to count the number of possible sequences of 4 aces either.
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  3. #3
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    Quote Originally Posted by mairjuanaman View Post
    52 cards, dealt 5 cards
    P(1st 4 are aces, last is face card)
    the answer is 1/1082900
    how do i do it (besides the [4/52][3/51][2/50][1/49][12/48])?
    shouldn't 4 nCr 4 x 12 nCr 1 all over 52 nCr 5 work?????!?!?!
    The order of the cards is important.
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  4. #4
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    52C4 = 270725

    Leaves 48/52 = 1/4

    1/270725 * 1/4 = 1/1082900
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  5. #5
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    Hello, mairjuanaman!

    52 cards, dealt 5 cards
    . . P(1st 4 are aces, last is face card)

    The answer is: . \frac{1}{1,\!082,\!900}

    How do i do it? . . . besides: . \frac{4}{52}\cdot\frac{3}{51}\cdot\frac{2}{50}\cdo  t\frac{1}{49}\cdot\frac{12}{48} . . This is correct.

    Shouldn't . \frac{(_4C_4)\cdot(_{12}C_1)}{_{52}C_5} . work? . . . . . no

    That is the probability of getting 4 Aces and 1 Face Card in any order.



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