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Math Help - Card Question

  1. #1
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    Card Question

    You randomly draw 1 card from a standard deck of 52, and 1 card from another standard deck. What's the probability that the sum of the 2 cards is 7?
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    You can solve it by solving this question :

    How many solution have to this linear equation:

    x_1 + x_2 = 7 , when x_1, x_2 are from the group: {1,2,3,4,5,6}

    ?
    Last edited by Also sprach Zarathustra; July 8th 2010 at 11:53 PM. Reason: 5=7 :)
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    Quote Originally Posted by lifeunderwater View Post
    You randomly draw 1 card from a standard deck of 52, and 1 card from another standard deck. What's the probability that the sum of the 2 cards is 7?
    First of all, what work have you done?

    Second, are aces counted as having value 1 or not?
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  4. #4
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    I thought of it that way >>> (not sure if I am correct though)
    since there are 2 standard deck of cards and the sum of 2 numbers goes from 1 through 6
    which means each card from 1-6 has 4 cards repeated. which makes 6*4= 24 and since
    we have 2 deck of standard cards of 52 then >>
    24/52+ 24/52 = 48/52= 12/13

    But I don't think that is correct?!?!
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by lifeunderwater View Post
    I thought of it that way >>> (not sure if I am correct though)
    since there are 2 standard deck of cards and the sum of 2 numbers goes from 1 through 6
    which means each card from 1-6 has 4 cards repeated. which makes 6*4= 24 and since
    we have 2 deck of standard cards of 52 then >>
    24/52+ 24/52 = 48/52= 12/13

    But I don't think that is correct?!?!

    Where you use the condition that the sum of the 2 cards is 7?
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    I don't understand what you asked above? could you please clarify. thank you!
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  7. #7
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    Quote Originally Posted by lifeunderwater View Post
    I thought of it that way >>> (not sure if I am correct though)
    since there are 2 standard deck of cards and the sum of 2 numbers goes from 1 through 6
    which means each card from 1-6 has 4 cards repeated. which makes 6*4= 24 and since
    we have 2 deck of standard cards of 52 then >>
    24/52+ 24/52 = 48/52= 12/13

    But I don't think that is correct?!?!
    Well you didn't answer my aces question so I'll just assume aces count as 1.

    We have 52^2 ways to choose the cards when order is taken into account.

    1 + 6 -> 4*4 = 16 ways

    2 + 5 -> 16 ways

    ...

    6 + 1 -> 16 ways

    we have \displaystyle P=\frac{6\cdot16}{52^2}.
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  8. #8
    MHF Contributor undefined's Avatar
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    By the way it can be seen that this is the same as the probability of getting a sum of 7 if we only had two 13-card decks with just one of each card value in them. We get 6/169.
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  9. #9
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    Quite different if 1 deck, and sum to be even; like:
    "You randomly draw 2 cards from a standard deck of 52.
    What's the probability that the sum of the 2 cards is 8?"

    1st card: 28/52 = 7/13

    2nd card: (6/7)(4/51) + (1/7)(3/51) = 9/119

    Probability: (7/13)(9/119) = 9/221

    Agree?
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  10. #10
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Wilmer View Post
    Quite different if 1 deck, and sum to be even; like:
    "You randomly draw 2 cards from a standard deck of 52.
    What's the probability that the sum of the 2 cards is 8?"

    1st card: 28/52 = 7/13

    2nd card: (6/7)(4/51) + (1/7)(3/51) = 9/119

    Probability: (7/13)(9/119) = 9/221

    Agree?
    Agreed; and just to be clear, that's drawing two cards from a single deck without replacement where aces count as 1.

    Brute force check in PARI/GP

    Code:
    Input: s=0;for(i=0,50,for(j=i+1,51,if(i%13+1+j%13+1==8,s++)));s
    Output: 54
    Input: 54/binomial(52,2)
    Output: 9/221
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  11. #11
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    Thanks, Mr n/0
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