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Probability Generating Functions Question

  1. Sep 29, 2015 #1
    1. The problem statement, all variables and given/known data
    In playing a certain game, your ability scores are determined by six independent rolls of three dice. After each set of six rolls, you are given the choice of keeping your scores or starting over.

    (a) How many times should you expect to start over in order to get a set of ability scores with at least two scores that are 18?

    2. Relevant equations
    Binomial Probability Formula = (N choose K)Pkqn-k

    E(X)=1/p
    From the PGF for Geometric Distribution

    3. The attempt at a solution

    Probability of rolling three sixes (18) is 1/216.

    P(Rolling 18 ≥ 2 in 6 trials) = 1 - P(0 18s) - P(1 18)
    = 1 - (215/216)6 - (6)(1/216)(215/216)5 ≈ .00031755
    (Binomial Probability Formula)

    Using the Geometric (.00031755) Distribution, E(X) = 1/p = 1/.00031755 ≈ 3149

    Where E(X) is the expected number of repeats before getting ≥ 2 18s.

     
  2. jcsd
  3. Sep 29, 2015 #2

    Ray Vickson

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    Please clarify: (1) If you decide to start over, do you discard your current scores and go back to 0? (It sound like you do.) (2) If on one roll of 6 dice you get a total of 36 , does that automatically count as two '18s'? (3) Are we allowed to make six rolls (of 6 dice in each roll), getting 6 each time, for a total of 36 = 18 ×2? (4) What has any of this to do with probability generating functions?
     
  4. Sep 29, 2015 #3
    If you start over, all of your current scores are discarded.

    The trials consist of 6 independent rolls of three dice. So, you have a set of 6 sums of the roll of three dice.
    So to get two "18s" you would need to roll three sixes twice in 6 attempts at rolling three dice.
     
  5. Sep 29, 2015 #4

    Ray Vickson

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    Of course: you did say that, more or less.
     
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