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Math Help - Dice problem

  1. #1
    Newbie TriForce's Avatar
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    Dice problem

    5 regular 6 sided dice are thrown. What is the probability to get at least 4 dice with 6:es?

    The "at least" part is a bit tricky.

    Solved it.

    Found this video that helped me solve it.

    Last edited by TriForce; March 6th 2014 at 04:18 AM.
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  2. #2
    MHF Contributor

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    Re: Dice problem

    Quote Originally Posted by TriForce View Post
    5 regular 6 sided dice are thrown. What is the probability to get at least 4 dice with 6:es?
    Do not over complicate it. Either four or five:
    $5\left( {\dfrac{5}{6}} \right){\left( {\dfrac{1}{6}} \right)^4} + {\left( {\dfrac{1}{6}} \right)^5}$
    Last edited by Plato; March 6th 2014 at 06:54 AM.
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  3. #3
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    Re: Dice problem

    "At least four" means "four or five". I don't see why that would be "tricky".

    Write "S" to mean a six, "N" to mean "not a six". Then one way of getting 4 sixes in five rolls would be "SSSSN", another would be "SSSNS", etc. It should be easy to see that there are five ways (NOT 4) to write those: the single "N" in each of the 5 places. And the probability of each of those is \left(\frac{5}{6}\right)\left(\frac{1}{6}\right)^4 so the probability of "four sixes in five rolls" is 5\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)^  4.

    The probability of 5 sixes in 5 rolls is, of course, \left(\frac{1}{6}\right)^5. The probability of "at least four sixes= either 4 sixes or 5 sixes" is the sum of those:
    5\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)^  4+ \left(\frac{1}{6}\right)^5.

    (Again, "5", not "4". That could also be calculated as \frac{5!}{4! 1!}.)
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  4. #4
    Newbie TriForce's Avatar
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    Re: Dice problem

    Yes, thanks for the clarifications, it's a lot easier to recognize the path to solution when the technique is clarified.

    And the technique is basically:
    First step is identifying where the at least bit is, in my case it was at 4 out of 5.
    Next step is to sum all remaining bits, including the staring one, with the following formula:
    P = n C r * p^r * q^(n-r) ~ as explained in the video.
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