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If I roll a die $100$ times, there are $6^{100}$ possible ordered outcomes.

Some of these outcomes, say $n$, will add to $347$, for example.

Is there a way to express $n$ in terms of Stirling numbers (or am I think of partitions?).

One way to estimate $n$:

  • note that the sum has a mean of $350$ with a standard deviation of $17.08$.

  • Use the normal distribution to calculate the probability that the sum is between $346.5$ and $347.5$.

  • Multiply this probability by $6^{100}$

Does this provide a new (or even useful) way of estimating Stirling numbers?

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    Stirling numbers enumerate partitions of a set into a fixed number of subsets or cycles of arbitrary size, and here we have a partition of a number $n$ into a sum of a fixed number of terms of bounded size... What kind of connection do you expect?2012-08-09

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