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Neither the player nor the dealer is dealt a blackjack

  1. Jul 2, 2007 #1
    I am trying to find the probability of neither the player nor the dealer being dealt a blackjack (an ace with a ten, jack, queen, or king) in a blackjack game with a freshly shuffled deck. This is what I did:

    1 - P(at least one gets a blackjack)

    = 1 - [ 4*16/(52c2) + 3*15/(50c2)]

    However, I get an answer that's about one percent higher than it should be. The right answer is .9052.
  2. jcsd
  3. Jul 3, 2007 #2
    According to conditional probability, we have P(~A and ~B) =P(~A)xP(~B|~A) Which I read, the probability of not A and not B is equal to the probability of not A times the probability of not B given not A)

    So in this case, we have not A = (1-(4*16/52C2) and not B given not A =1-64/50C2.

    If that is correct, then we have (.9478)(.9517) = .9020.

    HOWEVER, I see that this reasoning has a flaw in it, because it does not take into account that the first player may get, for example, an ace, but not have a blackjack.
    Last edited: Jul 4, 2007
  4. Jul 12, 2007 #3
    It turns out this problem has been solved by a group of actuaries. http://actuary.com/actuarial-discussion-forum/showthread.php?t=5022

    The solution is: P(AUB)=P(A)+P(B)-P(AB).

    In this case, P(A)=P(B), because both cases are independent and they both have equal odds. However, P(AB) = 4x16/52C2 * 3x15/50C2. Thus the probability of at least one blackjack is .0947579032. And the probability of no blackjack is .905242097.

    If this seems strange, a simpler example is: If two cards are dealt, what is the probability that at least one is red?

    P(AUB) = P(A) + P(B) -P(AB) = 1/2 + 1/2 -(26/52)(25/51) =.75490.

    This can be checked by looking at tree probability:

    1/2R to 25/51 R or 26/51 B

    And 1/2B to 26/51R or 25/51B.

    Thus if both are black we have (1/2)(25/51) =25/(102) = .24510; and at least one is red = 1-.24510= .75490.
  5. Jul 12, 2007 #4
    Yep, I posted the problem over there too...I was anxious to know the answer. I was going to post the solution here but forgot, thanks robert.
  6. Jul 14, 2007 #5
    We could look at this in the case of three players. First we consider just three cards and being dealt. Consider the color red:

    P(aUbUc) = P(a)+P(b)+P(c) - {P(ab)+P(ac) + P(bc)} + P(abc). (P(ab) means P(a intersection b)).

    But the probabilities in each group are the same, so this becomes:

    P(aUbUc) = 3P(a)-3P(ab)+P(abc)

    The P(a) = 1/2; P(ab)=(1/2)(25/51)=25/102; P(abc) = (1/2)(25/51)(24/50) =2/17.
    Thus P(aUbUc) = 3/2-75/102 + 2/17 = 15/17. While if all three cards are black, we independently find: (1/2)(25/51)(24/50) = 2/17.

    Applying the same method to the blackjack problem, we arrive at: The probability that at least one hand of three will have a blackjack is .139521.., and thus chance of no blackjack is approximately 86%.
    Last edited: Jul 14, 2007
  7. Sep 11, 2009 #6
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