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Math Help - Dice rolling problem

  1. #1
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    Dice rolling problem

    Hi

    A friend and I have been trying to figure out this seemingly simple problem for a while and couldn't find any help with it anywhere online. The problem is as follows:

    What are the odds of rolling a seven at least once (that is, once, twice OR three times) in three rolls of two dice?

    The closet we could come up with is (1/6)+(1/12)+(1/18) which gives the logical looking number of 30.5%, but (we think) that assumes you stop rolling if you roll a seven. At this point I can't even tell if that matters.



    (If you're wondering, the impetus for this question is the boardgame Settlers of Catan - I wanted to know, with three people playing, what the probability is of a 7 being rolled between the end of your turn and your next turn, i.e. 3 rolls)

    Thanks a lot.
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  2. #2
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    Hello, Zhankfor!

    What is the probabiity of rolling a "7" at least once in three rolls of two dice?
    For each roll, there are 6^2\,=\,36 outcomes.
    Six of them result in a "7": . P(7) \:=\:\frac{6}{36} \:=\:\frac{1}{7}
    Hence, the probability of not getting a 7 is: . P(\text{Other}) \:=\:\frac{5}{6}


    The opposite of "at least one 7" is "no 7's" ... that is, "3 Others"

    . . P(\text{3 Others}) \:=\:\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{5}{6} \:=\:\frac{125}{216}


    Therefore: . P(\text{at least one 7}) \;=\;1 - \frac{125}{216} \;=\;\frac{91}{216} \;\approx\;42\%

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  3. #3
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    Thanks a lot! I had at one point gotten to that basic idea (subtract the Others from all possibilities) but I couldn't figure out how to translate that into numbers.
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