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If I roll $m$ dice with $n$ sides and consider only the largest result, how much do I get on average?

In other words, if my formula is correct, is there a closed form for the sum $$\frac 1{n^m}\sum_{i=1}^ni(i^m-(i-1)^m)$$

  • for $m=1$, $(n+1)/2$
  • for $m=2$, $(n+1)(4n-1)/(6n)$
  • for $m=3$, $(n+1)(3n-1)/(4n)$
  • but in general?
mjqxxxx
  • 41,358

1 Answers1

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Your formula looks correct. To sum it, note that $$ \begin{eqnarray} \sum_{i=1}^{n}i\left(i^m - (i-1)^m\right) &=& \sum_{i=1}^{n}i^{m+1} - \sum_{i=1}^{n}i(i-1)^m \\ &=& \sum_{i=1}^{n}i^{m+1}-\sum_{i=1}^{n}(i-1)^{m+1}-\sum_{i=1}^{n}(i-1)^m \\ &=& \sum_{i=1}^{n}i^{m+1}-\sum_{i=0}^{n-1}i^{m+1}-\sum_{i=0}^{n-1}i^m \\ &=& n^{m+1}-\sum_{i=1}^{n-1}i^m, \end{eqnarray} $$ so $$ E_{m,n}=n-\sum_{i=1}^{n-1}\left(\frac{i}{n}\right)^m=n-\frac{1}{n^m}H_{n-1}^{(-m)}. $$ For large $n$, this is asymptotic to $n-n\int_{0}^{1}x^m dx=n\left(1-\frac{1}{m+1}\right).$ For finite $n$, there are additional correction terms, the first of which is always $+1/2$: $$ E_{m,n} = n\left(1-\frac{1}{m+1}\right)+\frac{1}{2}+O(1/n). $$

mjqxxxx
  • 41,358