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Math Help - Dice probability

  1. #1
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    Dice probability

    A dice is tossed 6 times. What is the probability that on each toss, the number shown is less than or equal to the previous tosses.
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  2. #2
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    I think my previous answer was wrong, let me do more research and then I'll post an answer unless somebody gets it first.
    Last edited by mathceleb; March 21st 2008 at 11:58 AM.
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  3. #3
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    Hello, math sucks!

    I have an approach to this problem.
    But I have no proof of my theory . . .


    A die is tossed 6 times. What is the probability that on each toss,
    the number shown is less than or equal to the previous tosses?

    Consider rolling two times.
    The outcomes are:

    . . \begin{array}{cccccc}<br />
{\color{blue}(1,1)} & (1,2) & (1,3) & (1,4) & 1,5) & (1,6) \\<br />
{\color{blue}(2,1)} &  {\color{blue}(2,2)} & (2,3) & (2,4) & (2,5) & (2,6) \\<br />
{\color{blue}(3,1)} & {\color{blue}(3,2)} & {\color{blue}(3,3)} & (3,4) & (3,5) & (3,6)\end{array}
    . . \begin{array}{cccccc}<br />
{\color{blue}(4,1)} & {\color{blue}(4,2)} & {\color{blue}(4,3)} & {\color{blue}(4,4)} & (4,5) & (4,6) \\<br />
{\color{blue}(5,1)} & {\color{blue}(5,2)} & {\color{blue}(5,3)} & {\color{blue}(5,4)} & {\color{blue}(5,5)} & (5,6) \\<br />
{\color{blue}(6,1)} & {\color{blue}(6,2)} & {\color{blue}(6,3)} & {\color{blue}(6,4)} & {\color{blue}(6,5)} & {\color{blue}(6,6)}\end{array}

    Out of the 6^2 outcomes,
    . . a triangular number, \frac{6\cdot7}{2} are desirable.


    Roll three times. .There are: . 6^3 outcomes.
    The outcomes will be displayed in a 6󬝲 cube.
    Picture a plane "bisecting" the cube through two opposite vertices.

    The desirable outcomes are in the the "larger half".
    This is a tetradedon with sides of 6.
    It contains: \frac{6\cdot7\cdot8}{3!} = 56 outcomes.


    Roll four times. .There are: . 6^4 outcomes.
    The ouctomes are displayed in a 6󬝲6 hypercube.
    The "larger half" contains: . \frac{6\cdot7\cdot8\cdot9}{4!} \:=\:126 outcomes.


    Let's skip to six rolls. .There are: . 6^6 outcomes.
    The outcomes are in a 6󬝲󬝲6 hypercube.
    The "larger half" contains: . \frac{6\cdot7\cdot8\cdot9\cdot10\cdot11}{6!} \;=\;462 outcomes.

    Therefore: . \text{Prob} \;=\;\frac{462}{46,656} \;=\;\frac{77}{7,776}

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  4. #4
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    Very interesting Soroban! Now we can generalize, if we toss the dice n times, the resulting probability would be
    \frac{\displaystyle\dbinom{n+5}{5}}{\displaystyle 6^n}
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