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Is there a general formula to determine the probability of unbounded, cumulative dice rolls hitting a specified number?

For Example, with a D6 and 14:

5 + 2 + 3 + 4 = 14 : success

1 + 1 + 1 + 6 + 5 + 4 = 17 : failure

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    The related question http://mathoverflow.net/questions/18282/probabilities-and-rolling-2-dice/18293#18293 at mathoverflow might be of interest. The consensus in that discussion was that there wasn't a "nice" explicit formula (although there is the partial-fraction formula that Moron gives below) but that these probabilities are easy to compute recursively.2010-11-07
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    I can brute force the probability up to n=10, and for large n the answer is 1/6. Can anyone brute force enough points for n=10...100 that a graphical representation of the results might indicate a good easy-to-calculate approximation of the correct answer?2010-11-07
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    @Sparr: the answer is 1/6 + O(r^n) for some r < 1. How interested are you in an upper bound on r?2010-11-07
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    (For what it's worth, r is about 0.73, but I am trying to estimate this number without the help of a machine.)2010-11-07
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    what is O() in that context?2010-11-07
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    @Sparr: Big-O notation. See http://rob-bell.net/2009/06/a-beginners-guide-to-big-o-notation/ and http://en.wikipedia.org/wiki/Big_O_notation2010-11-09
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    I am familiar with Big-O notation. O() in THAT context is not a function with a return value. Thus, how can "1/6 + O(r^n)" be an answer to my question?2010-11-10
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    @Sparr Checkout http://en.wikipedia.org/wiki/Big_O_notation#Infinitesimal_asymptotics specifically.2010-11-15
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    Ahh, ok, thanks. Unfortunately that only defines an upper limit, not a good approximation.2010-11-15

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