Suppose I roll an $n$-sided die once. Now you repeatedly roll the die until you roll a number at least as large as I rolled. What is the expected number of rolls you have to make?

I know the answer to this problem, but I'm curious about possible solutions people might post.


If you roll a $k$, then there are $n-k+1$ possible numbers out of $n$ that will be greater than or equal to $k$. This gives rise to a geometric distribution, and so the expected number of rolls required after rolling a $k$ is $\frac{n}{n-k+1}$. Averaging over all $k$, the expected number of rolls will be $$\mathbb{E}=\frac{1}{n}\sum_{k=1}^n \frac{n}{n-k+1}=H_{n}$$ where $H_n$ is the $n^{th}$ harmonic series.

  • $\begingroup$ Could you explain why the expected number of rolls required is n/n-k+1? $\endgroup$ – harekuin Apr 25 '17 at 3:18
  • $\begingroup$ This is a fundamental result about geometric distributions. If you have $k$ chance of success on a given roll, the expected number of rolls for success is $\frac 1k$ $\endgroup$ – Ross Millikan Apr 25 '17 at 3:22
  • $\begingroup$ I understand that (as in E[x] = 1/p for a geometric distribution.) then what is the the sum for? i dont fully understand the answer :(. why is the expected being multiplied by 1/n and then the summation?? $\endgroup$ – harekuin Apr 25 '17 at 3:28
  • $\begingroup$ The sum is to cover the range of first rolls. If the die has seven sides and the first roll is $4$, the chance on each roll to at least match it is $\frac 47$ so the expected number of rolls is $\frac 74$. There are seven possible first rolls and we sum over them. $\endgroup$ – Ross Millikan Apr 25 '17 at 3:34
  • $\begingroup$ oh i see. just a general question: the question asks for for the average roll we have to make. Do we NEED to do all possible k's? $\endgroup$ – harekuin Apr 25 '17 at 3:41

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