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The following is intuitive: if $p(n)$ is the probability of "rolling $n$ as the cumulative sum of arbitrarily many fair dice" then $p(n)\approx p(m)$ for $m$ and $n$ sufficiently large. This is proven in an answer here but I have a few questions:

  1. What exactly has been proven? In other words, how can I formulate this probability more precisely?
  2. How can we see that the above claim is true without explicitly computing the distribution? It seems like this should follow from some sort of law of large numbers or result about the limiting distribution of the sum of i.i.d. random variables, but I can't see it.
  3. Is there a way to see easily, i.e. without too much computation, that the limiting probability has to be the inverse of the expectation of one roll?
user293794
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  • Perhaps you'll find the answers here more intuitive. – lulu Jul 10 '20 at 00:56
  • I showed here, but with a good deal of computation, that if you roll an $s$-sided die, numbered from $1$ to $s$, that the probability goes to $\frac2{s+1}$ but it didn't strike me that that was the reciprocal of the expectation of a single roll. I'd thought about the case where the numbers could be arbitrary, but it seemed too difficult, at least by the route I took. Now I'll try again. Thanks. – saulspatz Jul 10 '20 at 02:37

1 Answers1

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  1. What is proven in the linked answer is the following. Fix $N$. Uniformly randomly pick $k$ with $1 \leq k \leq N$, throw $k$ fair independent 6-sided dice, and add up the number of pips showing. Let $p(N)$ be the probability that this procedure results in $N$ pips. Then $\lim_{N \to \infty} p(N) = 2/7$. Note that it makes sense to restrict $k \leq N$ since there must be at least $k$ pips, so $k > N$ would contribute $0$.

  2. See Arthur's comment, "If you throw a die a lot of times and add your results as you go along, the sum increases by 3.5 on average for each roll, so my gut says that your probability converges to 1/3.5≈0.286 as $N$ grows." I don't find this particularly intuitive myself, but to each their own.

  3. From my perspective, the linked answer is not too computation-heavy. You don't really need the computer to tell you where the roots are, and the calculation $\lim_{x \to 1} G(x) (1-x) = 2/7$ [note that there was a typo] can be done without effort by noting $q(x) = \frac{x}{6} \frac{1-x^6}{1-x}$.

Joshua P. Swanson
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