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    $\begingroup$ So the point is that any problem like this must be compared to $\log \log N$. But to get any good estimates of the comparison is extremely hard, because $\log\log N$ is so much smaller than $\infty$. For example, in the problem above, based on $N = 10^9$, we could argue that in fact primes very much do like to divide the sum of their predecessors, as there are in fact 5 primes with this property, which is fifty percent more than the expected number. $\endgroup$ Commented Feb 1, 2013 at 20:18
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    $\begingroup$ Why is the probability $1/p$? $\endgroup$ Commented Feb 2, 2013 at 0:52
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    $\begingroup$ @Filippo Alberto Edoardo: It is a heuristic (as said in the answer) but one that in this cases seems to make perfect sense: the sum of the primes preceeding $p$ is a number significantly larger than $p$ (about size $p^2/ (2 \log p)$ , I think) and there seems no reason for it being biased towards being in any particular residue class moduo $p$; so that it is $0$ mod $p$ seems just as likely as it being anything else, so prop. $1/p$ for it being $0$, ie divisible by $p$. This is the natural heuristic in this case, and explains the scarcity. To make this heuristic precise, seems very hard. $\endgroup$ Commented Feb 2, 2013 at 12:48
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    $\begingroup$ Thanks, I did not intepret the $1/p$ as being the sentence "lie in one of the $p$ classes $\mod{p}$". $\endgroup$ Commented Jan 22, 2014 at 23:53
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    $\begingroup$ (Nitpicking: The guy's name was Mertens, not Merten.) $\endgroup$ Commented Feb 19, 2014 at 11:25