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Math Help - Dice probability problem

  1. #1
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    Dice probability problem

    I have this question: What is the probability that 3 100-sided dice would exceed 222?

    Well, I thought I could use joint pdf and a uniform distribution to solve this problem.

    so I had
     <br />
\int_{222}^{300} \int_{222-x}^{300} \int_{222-y-z}^{300} 1/100^3\,dx\,dy\,dz<br />


    But it looks like this doesn't work.

    Any help would be appreciated.
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  2. #2
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    Re: Dice probability problem

    Hello, ChaosticMoon!

    What is the probability that three 100-sided dice would exceed 222?

    I assume each die has sides numbered from 1 to 100.

    There are 100^3\,=\,1,\!000,\!000 possible outcomes.

    Visualize a cube graphed in the first octant of an xyz-system.
    One vertex is at the origin; the opposite vertex is at (100, 100, 100).

    Consider all the lattice points (those with integer coordinates)
    . . from (1,1,1) to (100,100,100).
    These represent the 1,000,000 possible outcomes.

    The points whose coordinates have a sum exceeding 222
    . . are "outside" the triangle with coordinates:
    . . (22, 100, 100), (100, 22, 100), (100, 100, 22).

    How many lattice points are contained in this tetrahedron?

    The tetrahedron has 21 "levels".
    Each level contains a triangular number of points.

    We want the sum of the first 21 triangular numbers.

    Fortunately, there is a formula for this: . N \:=\:\frac{n(n+1)(n+2)}{6}
    For n = 21\!:\;\;N \:=\:\frac{(21)(22)(23)}{6} \:=\:1771

    Therefore: . P(\text{sum}> 222) \:=\:\frac{1,\!771}{1,\!000,\!000}

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  3. #3
    Grand Panjandrum
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    Re: Dice probability problem

    Quote Originally Posted by Soroban View Post
    Hello, ChaosticMoon!


    I assume each die has sides numbered from 1 to 100.

    There are 100^3\,=\,1,\!000,\!000 possible outcomes.

    Visualize a cube graphed in the first octant of an xyz-system.
    One vertex is at the origin; the opposite vertex is at (100, 100, 100).

    Consider all the lattice points (those with integer coordinates)
    . . from (1,1,1) to (100,100,100).
    These represent the 1,000,000 possible outcomes.

    The points whose coordinates have a sum exceeding 222
    . . are "outside" the triangle with coordinates:
    . . (22, 100, 100), (100, 22, 100), (100, 100, 22).

    How many lattice points are contained in this tetrahedron?

    The tetrahedron has 21 "levels".
    Each level contains a triangular number of points.

    We want the sum of the first 21 triangular numbers.

    Fortunately, there is a formula for this: . N \:=\:\frac{n(n+1)(n+2)}{6}
    For n = 21\!:\;\;N \:=\:\frac{(21)(22)(23)}{6} \:=\:1771

    Therefore: . P(\text{sum}> 222) \:=\:\frac{1,\!771}{1,\!000,\!000}
    But the answer is ~8%

    (the normal approximation is sort of appropriate for the sum of three dice, it gives an answer off by about 0.4%points)

    CB
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