# Find the closed form of an n-sum

I'd like to find the closed form or a quickly converging rewriting of the following n-sum:

$$\left(\frac{1}{6}\right)^n\sum_{x_1=1}^\infty\sum_{x_2=1}^\infty\ldots\sum_{x_n=1}^\infty\left(\frac{5}{6}\right)^{x_1+x_2+\ldots+x_n-n}*max(x_1,x_2,\ldots,x_n)$$

The problem this arises from is when you have n 6-sided dice, roll all of them, set aside the 6's, roll the rest, set aside 6's again and so on until all dice faces are 6's. These sums are the expectation value for the number of rolling rounds to get n 6's.

By numerical summation I got $\approx8.727$ for $n=2$ and $\approx10.555$ for $n=3$ but it's getting impossible to calculate for larger n's and it converges very slowly because of the long tail.

Thanks,

Marton

• Is your sum the same as$$\sum_{r=0}^\infty(1-(1-(\frac56)^r)^n)?$$ – bof Oct 1 '14 at 9:06
• Based on the n=2,3 values it is the same, how did you get it? – Marton Oct 1 '14 at 9:17
• The term $1-(1-(\frac56)^r)^n$ is the probability that your goal of all 6's has not been achieved in the first $r$ rounds, so another round is needed. I was thinking of posting this as an answer, but there's no need, because I see that Did has updated his answer to include this formula. – bof Oct 1 '14 at 9:53

Standard generating functions technology yields that the $n$th sum $s_n$ is such that, when $n\to\infty$, $$\lim_{n\to\infty}\frac{s_n}{\log n}=\frac1{\log(6/5)}\approx5.485.$$ Is this the kind of result you are after?
For small values of $n$, one could use the exact value as a finite sum in the RHS of the identities $$s_n=\sum_{i\geqslant0}\left(1-(1-(5/6)^i)^n\right)=\sum_{k=1}^n{n\choose k}(-1)^{k+1}\frac{1}{1-(5/6)^k}.$$ For example, $s_1=6$, $s_2=\frac{96}{11}\approx8.73$, $s_3=\frac{10566}{1001}\approx10.56$, as you mentioned, and $s_4=\frac{728256}{61061}\approx11.93$, $s_5=\frac{3698650986}{283994711}\approx13.02$, $s_6\approx13.94$, $s_7\approx14.72$, and so on...
For middle-sized values of $n$, a more eficient formula would be $$s_n=1+\sum_{k=1}^n{n\choose k}(-1)^{k+1}\frac{5^k}{6^k-5^k}.$$
• For every nonnegative integer valued random variable $T$, $$E(T)=\sum_{i\geqslant0}P(T\gt i).$$ Here, $[T\leqslant i]$ means that $i$ draws were sufficient to get a six on each of the $n$ dice. Which does not happen on a given die with probability $p^i$, hence it happens on every dice with probability $P(T\leqslant i)=(1-p^i)^n$ with $p$ equal to what you guess... et voilà! – Did Oct 1 '14 at 10:00