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Math Help - probability of a deck ofcards

  1. #1
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    probability of a deck ofcards

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    Last edited by badboychow; July 12th 2007 at 09:33 AM.
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  2. #2
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    Quote Originally Posted by badboychow View Post
    You draw 2 cards from a deck of 52 cards and the return as follow:

    Result
    How much you win
    Blackjack or Double Aces
    $50

    Total is more than 16
    $10

    None of the above
    - $5

    (Assume the face value of an ace is 11)

    Blackjack or double aces

    [(16/52) x (4/51)] + [(4/52) x (3/51)]
    *is this correct?

    I'm not sure of the rest. Anybody help pls
    Blackjack or Double Aces.

    first card is a "10" second an ace (16/52)x(4/51)

    first card is an ace second card an ace or "10" (4/52)x(19/51)

    so probability is: (16/52)x(4/51) + (4/52)x(19/51)

    RonL
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  3. #3
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    thanks but how about the probability where total is more than 16 and none of the above...i have no idea...
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  4. #4
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    Let's see if we can pick it apart.

    The probability of drawing two cards from a deck and their sum is > 16.

    There's a trick here because of Jack, Queen, King also count as 10.

    We can have Ace-Ace and Ace-not Ace. To sum over 16 we must have A6, A7, A8, A9, A10, AJ, AQ, AK, or AA. \underbrace{(\frac{4}{52})}_{\text{Ace}}\overbrace  {(\frac{32}{48})}^{\text{not Ace}}+\underbrace{(\frac{4}{52})}_{\text{Ace}}\ove  rbrace{(\frac{3}{51})}^{\text{Ace}}=\frac{37}{663}

    For 10, J, Q, K we have 4\left[(\frac{4}{52})(\frac{28}{48})+(\frac{4}{52})(\frac  {3}{51})\right]=\frac{131}{663}

    For 9, we have (\frac{4}{52})(\frac{24}{52})+(\frac{4}{52})(\frac  {20}{48})=\frac{19}{442}

    For 8, 7, and 6 we have (\frac{4}{52})(\frac{24}{48})+(\frac{4}{52})(\frac  {20}{48})+(\frac{4}{52})(\frac{4}{48})=\frac{1}{13  }

    Add them up and get \frac{165}{442}

    I could easily be over or under counting.
    Last edited by galactus; July 11th 2007 at 02:46 PM.
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  5. #5
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    thanks thanks....now i playing cards...seeing how many probability....haha
    Last edited by badboychow; July 12th 2007 at 09:35 AM.
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  6. #6
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    Quote Originally Posted by badboychow View Post
    You draw 2 cards from a deck of 52 cards and the return as follow:
    Result
    How much you win
    Blackjack or Double Aces
    $50
    Total is more than 16
    $10
    total more than 16 but not blackjack or double ace:

    1st card A 2nd card 6, 7, 8, 9 (4/52)x(4/51)

    1st card "10" 2nd card 7, 8, 9, "10" (4/52)x(27/51)

    1st card 9 2nd card 8, 9, "10", A (4/52)x(27/51)

    1st card 8 2nd card 9, "10", A (4/52)x(24/51)

    1st card 7 2nd card "10" A (4/52)x(20/51)

    1st card 6 2nd card A (4/52)x(5/51)

    so probability is:

    (4/52)x(4/51) + (4/52)x(27/51) + (4/52)x(27/51) + (4/52)x(24/51) + (4/52)x(20/51) + (4/52)x(5/51)

    RonL
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  7. #7
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    Hello, badboychow!

    You draw 2 cards from a deck of 52 cards.

    . . \begin{array}{cc}\text{Blackjack or Double Aces:} & \text{win \$50} \\<br />
\text{Total is more than 16:} & \text{win \$10} \\<br />
\text{Otherwise:} & \text{lose \$5}\end{array}

    (Assume the face value of an Ace is 11)

    Find the expected value.
    There are: . {52\choose2} = 1326\text{ possible pairs.}

    Double Aces: .There are {4\choose2} = 6 ways.
    Blackjack: .There are {4\choose1} = 4 ways to get an Ace.
    . . . . . and {16\choose1} = 16 ways to get a card worth 10.
    . . So there are: . 4\cdot16 \:=\:64 Blackjacks.

    Hence, there are: . 6 + 64 \:= 70 ways to win $50.


    Sum greater than 16

    We can have an Ace (4 choices) and any of 6, 7, 8, or 9 (16 choices).
    . . There are: . 4\cdot16 \,=\,64 ways.

    We can have a 7 (4 ways) and any "10" (16 ways).
    . . There are: . 4\cdot16\,=\,64 ways.

    We can have an 8 (4 ways) and a 9 or a "10" (20 ways).
    . . There are" . 4\cdot20 \,=\,80 ways.

    We can have two 9's. .There are: . {4\choose2} = 6 ways.

    We can have a 9 (4 ways) and a "10" (16 ways).
    . . There are: . 4\cdot16 \,=\,64 ways.

    We can have two "10's". .There are: . {16\choose2} = 120 ways.

    Hence, there are: . 64 + 64 + 80 + 6 + 64+ 120 \,= 398 ways to win $10.


    Then there are: . 1326 - 70 - 398 \,= 858 ways to lose $5.


    The expected value is: . E \;=\;(50)\left(\frac{70}{1326}\right) + (10)\left(\frac{398}{1326}\right) + (-5)\left(\frac{858}{1326}\right)

    . . E \;=\;\frac{3190}{1326}\;=\;2.405731523


    Therefore, we can expect to win an average of about $2.41 per game.

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  8. #8
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    thank you so much....i think i know how to solve it....
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