# What's the probability that a sum of dice is prime?

Prompted by today's Minute Math question on the MAA site (http://amc.maa.org/mathclub/5-0,problems/T-problems/T-web,ia/2005web/tb05-12-ia.shtml), I started thinking about the probability that the sum of the numbers rolled on a set of $n$ dice is prime, particularly the asymptotics as $n\rightarrow\infty$. Heuristics strongly suggest that this is proportional to $1/\mathrm{ln}\ n$, and in fact, that it's $1/\mathrm{ln}\ n - O(1/\mathrm{ln}^2n)$, but I'm wondering how I would go about getting better asymptotics on the second term.

For the record, the heuristic argument goes something like this: assume for concreteness' sake that we're rolling 6-sided dice. Then the sum of the dice is closely approximated by a normal variable with mean $\mu=7n/2$ and variance $\sigma^2=35n/12$, and since the PNT says that the 'probability' of an integer $n$ being prime is roughly $1/\mathrm{ln}\ n$, we should be able to integrate that probability with respect to the normal distribution: $$p = {1\over \sqrt{2\pi\sigma^2}}\int_n^{6n} e^{-\left({(t-\mu)^2\over 2\sigma^2}\right)} {1\over\mathrm{ln}\ t} dt$$ And since $1/\mathrm{ln}\ t$ is monotonic, the value of the integral is bounded by the values we get by replacing its term in the integral with its maximum and minimum values on the integration interval: $${1\over\mathrm{ln}\ 6n} {1\over \sqrt{2\pi\sigma^2}}\int_n^{6n} e^{-\left({(t-\mu)^2\over 2\sigma^2}\right)} dt < p < {1\over\mathrm{ln}\ n} {1\over \sqrt{2\pi\sigma^2}}\int_n^{6n} e^{-\left({(t-\mu)^2\over 2\sigma^2}\right)} dt$$ Both of the integrals in the latter formula are essentially 1 (by definition), so we get ${1\over\mathrm{ln}\ 6n} < p < {1\over\mathrm{ln}\ n}$; replacing $\mathrm{ln}\ 6n$ by $(\mathrm{ln}\ 6 + \mathrm{ln}\ n)$ and using the binomial formula gives the heuristic approximation I alluded to above. This leads me to a couple of questions:

1. How safe is the heuristic argument above? I know that the PNT gives good bounds on the number of primes in an interval (on the order of $n^{1/2}$ here, which in particular means that the error from the prime-counting would be $O(n^{-1/2})$ and so much smaller than the inverse-log terms above), but my analytic number theory isn't good enough to know whether 'weighting' by the normal distribution would throw off the classical proofs.
2. How would I go about evaluating the integral above? Obviously the bounds I use bring it in to a fairly small range, but it seems as though to get a second term in my asymptotics I'd need to be able to at least approximate the integral, and there aren't any obvious tricks that look like they'd handle it well...
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You have noted this in your comment to an answer, but you should be aiming for probabilities near the centre of the distribution so $1/\ln(7n/2) = \dfrac{1}{\ln(n)+\ln(3.5)}$ – Henry Mar 8 '12 at 11:12

For (1.), Rosser and Schoenfeld (sp?) non-asymptotic estimates bound $\pi(n)$ between functions of the form $x/(\log x + C)$ and this should be enough.
For (2) your integrals are rapid decaying and incredibly close to the integrals on the whole real line. O($n$) standard deviations from the average is quite an unlikely event.