> Exactly. I knew that 1/p number didn't look right. Isn't it more like
> 1/sumof(all p < current p)?
The method that I used (I think I got it from _Primes and Programming_ by
Peter Glibin) is that for any given set of primes, the probability that any
number greater than them will be divisable by at least one of them is
~1-prod((p_i-1)/p_i).
The probability that any given prime p will divide another (>=2*p) is 1/p.
Therefore, the probability that it will not be divisable is (p-1)/p.
Now to get the probability that the number will not be divisable by two primes
p_1 and p_2, we multiply the individual probabilities that they do not divide
=(p_1-1)*(p_2-1)/(p_1*p_2). Therefore to find the probaility that they do
divide (since prob. of divisability +prob of nondivisability=1), we take
1-(prob that they do not divide)=1-(p_1-1)*(p_2-1)/(p_1*p_2), and so on
for p_3,p_4....p_n. Note that this primes do not have to be in numerical order
after a time, for suficiently many primes the only ones missed will be powers
of the primes not included which are very insignificant statistically.
We can also use this method as a slight alteration of Euclid's proof for an
infinity of primes, which I leave as an excersize for the readers ;)
-Lucas Wiman
P.S. This is not rigorously proven, and might not take into account certain
things like numbers divisable by p_1*p_2*p_3 or something like that.
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