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Math Help - blackjack proabability question

  1. #1
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    blackjack proabability question

    suppose that you are playing blackjack against a dealer. in freshly shuffled deck, what is the probability that neither you nor the dealer is dealt a blackjack?
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  2. #2
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    a blackjack is when you are dealt an ace and either a ten, jack, queen, or king.
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  3. #3
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    I think I have an answer although I always question my reasoning.

    Let A be the event - player gets a blackjack

    Let B be the event - dealer gets blackjack

    P(A) = (16*4)/(52 choose 2)

    By symmetry P(B) is the same.

    We have P(A or B) = P(A) + P(B) - P(A and B)

    So if we calculate P(A and B) we can subtract P(A or B) from 1 to get the answer.

    This is where I am unsure of myself.

    The total number of possible outcomes is (52 choose 2)*(50 choose 2).

    We give the first player 2 cards from 52 and the second 2 cards from the remaining 50.

    The number of outcomes where both the player and dealer have blackjack is 16*15*4*3.

    So we have P(A and B) = (16*15*4*3)/((52 choose 2)*(50 choose 2))

    I think this gives the right answer.
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  4. #4
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    by the way, the answer in the back of my book is .9052. what i just got is reasonably close to that. but i still question how i got the answer.
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  5. #5
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    Think carefully about what the question asks!
    What is the probability that neither you nor the dealer is dealt a blackjack?

    That is neither event happens: P\left( {\overline A  \cap \overline B } \right) = P\left( {\overline {A \cup B} } \right) = 1 - P\left( {A \cup B} \right).

    You dont get a blackjack and the dealer does not get a blackjack.

    Now use your answer for P\left( {A \cup B} \right) to finish.
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  6. #6
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    yes, i subtract it from 1. but was my combinatorial reasoning correct in coming up with P(A and B)?
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  7. #7
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    Quote Originally Posted by kkoutsothodoros View Post
    was my combinatorial reasoning correct in coming up with P(A and B)?
    Yes that is what I got.
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