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Rolling Dice and Probability

  1. Mar 29, 2012 #1
    1. The problem statement, all variables and given/known data

    Screen_shot_2012_03_29_at_8_55_48_AM.png

    3. The attempt at a solution
    The Fundamental Rule states that, given the six choices of value on a die, the number of possible outcomes for rolling three dice is: [itex](6)(6)(6) = 6^3 = 216[/itex]. Of these 216 possible outcomes, however, there is only one event that has a sequence of three sixes. Thus, given a single roll of the dice, the probability of not seeing a sequence of three sixes is [itex]\frac{215}{216}[/itex]. When the dice are rolled n-number of times, the probability of observing at least one sequence of triple sixes can be calculated as one minus the probability of no sequences of three sixes. We can write,

    [itex] 1 - ( \frac{215}{216})^n = f(n)[/itex].

    Let n be 149, and [itex]f(149) = .4991[/itex]. Let n be 150, and [itex]f(150) = .5015[/itex]. Thus, I conclude that 150 is the minimum n for a favorable outcome.

    Is that right?
     
  2. jcsd
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