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Math Help - Dice and Probability

  1. #1
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    Dice and Probability

    Okay, I have a 10-sided die. I'm looking for occurrences of 1's and 6's when I roll it.

    On a single role, the chance that I'll get a 1 or a 6 is 20%.

    I know that much.

    But lets say I don't get it on the first roll. Or the first four?

    Here's what I'm trying to figure out:

    Probability says that if I roll a d10, there's a 1 in 5 chance that I'll get a 1 or a 6 as a result. If I keep rolling, I'm GOING to get a 1 or a 6 eventually.

    What are the odds that it'll happen within the first four tries?
    What are the odds that two of the results will be 1 or 6?
    Three of the results?
    All four?
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  2. #2
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    Quote Originally Posted by mpen View Post

    Probability says that if I roll a d10, there's a 1 in 5 chance that I'll get a 1 or a 6 as a result. If I keep rolling, I'm GOING to get a 1 or a 6 eventually.

    What are the odds that it'll happen within the first four tries?
    For X is the probaility of rolling a 1 or a 6. make p = 0.2 and n = 4

    X is binomial, do you know this distribution?
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  3. #3
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    I have no idea what you just said. You had me until after the end of the first sentence.
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  4. #4
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    Quote Originally Posted by mpen View Post
    I have no idea what you just said. You had me until after the end of the first sentence.
    Well I have no idea what you are asking!
    There seems to be a 10-sided die.
    But you tell us nothing about how the sides are numbered.
    Are there only integers 1 to 6? Or 1 to 10?
    Try to clear up the ambiguities in your question.
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  5. #5
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    Quote Originally Posted by mpen View Post

    What are the odds that it'll happen within the first four tries?
    Well instead of using the word 'odds' the question should be 'What is the probability that it'll ...'

    If it happens once, twice, 3 or 4 times in the four tries it has happened, so you need to find P(X\geq 1)  = P(X=1)+P(X=2)+P(X=3)+P(X=4)

    Where P(X= x) = ^nC_x\times p^x\times (1-p)^{n-x}
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  6. #6
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    It's a d10. Like from a tabletop game. A ten-sided die with the numbers 1-10 on it.

    The probability of rolling a 1 or a 6 is 1/5. I can get that far. But I don't know how subsequent rolls affects that probability. If I roll twice, does that mean I have a 40% chance of getting 1 or 6 instead of just 20%?

    I'd like to know the chances that if I roll four times, I'll get one result that is either 1 or 6. And then two of the four results. Three and four after that.

    EDIT: This isn't for any class or academic purpose. I'm just genuinely curious about it.
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  7. #7
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    Quote Originally Posted by pickslides View Post
    Well instead of using the word 'odds' the question should be 'What is the probability that it'll ...'

    If it happens once, twice, 3 or 4 times in the four tries it has happened, so you need to find P(X\geq 1)  = P(X=1)+P(X=2)+P(X=3)+P(X=4)

    Where P(X= x) = ^nC_x\times p^x\times (1-p)^{n-x}
    I probably should have specified, but it's been 6 years since I opened a math book or sat in front of a math teacher. Those formulas are almost gibberish to me. Especially that second one. I don't know how to solve them.
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  8. #8
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    Quote Originally Posted by mpen View Post
    I probably should have specified, but it's been 6 years since I opened a math book or sat in front of a math teacher. Those formulas are almost gibberish to me. Especially that second one. I don't know how to solve them.
    Start by reading this Binomial probability - Wikipedia, the free encyclopedia
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  9. #9
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    I really appreciate the link, and that you're trying to help me learn it... but I'm really bad at math. The entire reply and the linked article make no sense to me. I won't be able to solve this by myself.
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