Rolling $c$ consecutive identical values of $n$-sided die

Asked Jun 23 '23 at 21:17

Active Jun 24 '23 at 21:33

Viewed 112 times

My question below is motivated by the following question (generalized from “The Riddler” blog on FiveThirtyEight website): What is the expected number of rolls to get $c\ge1$ consecutive identical values of an $n$-sided die ($n\ge1$)? It's a cute problem, the answer is $E_{n,c}=\sum_{i=0}^{c-1}n^i$ (exercise!). Clarification: This concerns $c$ in a row of any non-predetermined value.

OK so this becomes interesting with the following curious result: The expected number of coin flips to get three-in-a-row ($n=2$, $c=3$) is equal to the expected number of usual-die rolls to get two-in-a-row ($n=6$, $c=2$), namely 7, yet it's still more likely for the die-rolls to successfully finish before the coin-flips (due to different probability distributions).

My question is about the general scenario: What are the set of pairs $(n,c)$ which have the same answer $1+n+\cdots+n^{c-1}$?

We have $(n_1^{c_1}-1)(n_2-1)=(n_2^{c_2}-1)(n_1-1)$ for starters. We may as well ignore the pairs $(1,c)$ and $(n,2)$ as they get appended to every set as $c$ and $n$ vary, and relatedly $\lbrace(n,1)\;|\;n\in\mathbb N\rbrace$ is the set of answer 1.

edited Jun 24 '23 at 21:33

asked Jun 23 '23 at 21:17

Chris Gerig

2,151

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$$\sum_{i=0}^{c-1} n^i = \frac{n^c-1}{n-1}$$ – aschepler Jun 23 '23 at 21:20
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Are you really interested in solutions with $n=1$? – Ron Kaminsky Jun 23 '23 at 21:28
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I would have thought $\sum\limits_{i=1}^{c}n^i$ was more plausible (so $n$ times as many) - see the related https://math.stackexchange.com/a/597597/6460 – Henry Jun 23 '23 at 21:29
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And you probably mean "set of pairs of pairs", no? – Ron Kaminsky Jun 23 '23 at 21:31
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@Henry: Presumably the factor of $n$ is that you don't care which value gets repeated? – Brian Tung Jun 23 '23 at 21:41
Yes, note that the motivational question is not of the sort "number of flips to get $c$ heads in a row". – Chris Gerig Jun 23 '23 at 21:44
Allowing $c=2$ also isn't very interesting. You should edit the question to add all these qualifications. – Ron Kaminsky Jun 23 '23 at 21:59
@BrianTung - OK - seeing this in 538's Riddler means the you can indeed remove that factor of $n$ – Henry Jun 23 '23 at 23:18
A non-simple example is $(2,5)$ and $(5,3)$ both giving $31$, as does the simple $(30,2)$ and the trivial $(1,31)$ – Henry Jun 23 '23 at 23:56

Rolling $c$ consecutive identical values of $n$-sided die

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