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Probability quesiton

  1. Dec 16, 2004 #1
    ok, so what is the probability that the sum of 8 will appear before the sum of 7 when rolling 2 fair dices. How do I even start this?

    so with 2 fair dice, the probabality of gettng a sum of 8 is 5/36 right(2 6, 3 5, 4 4, 5, 3, 6 2)? and I got the same for the sum of 7(1 6, 2 5, 3 4, 4 3, 5 2, 6 1), 5/36.

    so what do I do next?
     
  2. jcsd
  3. Dec 17, 2004 #2

    Astronuc

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    Staff: Mentor

    I think you meant there are 6 ways to get a sum of 7:

    (1 6, 2 5, 3 4, 4 3, 5 2, 6 1) or 6/36 so there is a slightly greater probability of getting a 7 instead of 8 (only 5 combinations, or 5/36 of chance).

    If they were equal, the probability of getting a 7 first would be 0.5.
     
  4. Dec 17, 2004 #3

    shmoe

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    Break it up according to when the 8 appears. Can you find the probability that an 8 is thrown on toss n, but neither a 7 nor an 8 was thrown before that? How will this help?
     
  5. Dec 18, 2004 #4
    That's making the question more complicated than it is. You just imagine rolling the die until the first 7 or 8. So you have a given: the last roll of this sequence is either a 7 or an 8. If it's a 7 then you rolled a 7 before rolling an 8 (imagine for a moment extending the sequence beyond the 7 until the first 8 to be sure of this); if it's an 8 then you rolled an 8 before rolling a 7. The conclusion is easy.
     
  6. Dec 18, 2004 #5

    Astronuc

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    There are 6 combinations that yield 7, and 5 combinations of getting 8.

    All other combinations are irrelevant to this problem, because you are only interested in the total of 11 combinations that give 7 or 8.

    So the probability of getting a combination that gives 7 = 6/(6+5) and the probability of getting 8 is 5/(6+5).

    So what's the answer?
     
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