20090827, 19:30  #1 
2×13^{2}×17 Posts 
Odds that a random number is prime
A random 321,000 (decimal) digit number is chosen. What is the probability that it's prime if it...
a.) Is an odd number? b.) Doesn't have any factors below one billion (10^9)? c.) Doesn't have any factors below one trillion (10^12)? For part A, I got 1 in 370,000 by taking ln(10^321000) and dividing it by 2. Is this right? Also, could you help me out with parts B and C? 
20090828, 00:26  #2 
"William"
May 2003
New Haven
23×103 Posts 
http://en.wikipedia.org/wiki/Mertens%27_theorems
The third theorem is usually rearranged to get an estimate. Last fiddled with by wblipp on 20090828 at 00:26 
20090828, 11:26  #3  
Nov 2003
2^{2}×5×373 Posts 
Quote:
(b) approx. exp(gamma)/[9 * log(10)] (c) approx. exp(gamma)/[12*log(10)] Where pi is the prime counting function and gamma is Euler's constant. Look up Mertens' Theorem. Your answer for (a) is approximately correct. 

20090828, 16:50  #4 
2×3×17×31 Posts 
 In primenumbers@yahoogroups.com, James Wanless <james@...> wrote:
Hello All, I am interested in trying to derive (an estimate of) an expression for the likelihood [expressed as a probability between 0 and 1] that a given generic number (eg random) N is prime, given that no factors of it have yet been found of a size up to some lesser integer, M. I found this expression, which makes use of Merten's Theorem: http://home.earthlink.net/~elevensmo....html#Residual However, on first seeing Merten's Theorem I (mistakenly?) assumed that this _in itself_ [ie in entirety] was the answer I needed above, rather than the somewhat more complex answer in the link. I also (mistakenly?) had guestimated to myself (on very first thoughts) that my required expression might actually be independent of N [so long as within certain limits, eg M<=sqrt(N)]. Note that this is pretty much the case with pure Merten's. Surely, therefore, what I am saying is that the expression in the link is too simplistic [or rather actually overcomplex], because, for instance, it does not take into account the fact that not all primes in its population under consideration are equally likely  in fact the larger ones will be much rarer, probabilitywise. Am I going nuts here, or can someone else see what I'm getting at? [obviously  for my nefarious purposes :)  ie gaining some evidence that MM127 is prime :)  I'm looking for a probability answer as high as possible, but I do genuinely believe atm that the currently accepted estimate is too low...] Please help! James Firstly, thanks for your input regarding this, folks... I too have been thinking about it a bit more, and this, I think is my query/problem/objection: ******************************************************************** Suppose you give me a supposedly 'random' integer N. Suppose I then discover (empirically) that it is not divisible by any number (or indeed prime number) less than say 1000000 [ie M=1000000] Am I not entitled to be surprised  _regardless_ of the size of number N you originally gave me? ******************************************************************** ie  just a 'pure' Mertens result  rather than the lowering adjustment to Mertens... (IMHO) Probability is _so_ tricky, so I'm not claiming infallibility here, by any means  but I'd really appreciate it if someone could clear this up for me! I _think_ I might finally have an answer to this (after much consideration...)?? [Mertens' theorem is a redherring to this situation] ********************************************* pr(N prime  no factors up to M) = ln (M^2) / ln N ********************************************* J ...which would mean that the 'surprise factor' to discover that N has no factors up to M _would_ be independent of N, and just: ****************************** ln (M^2) ****************************** Now how exactly one interprets that 'surprise factor'...? J 
20090828, 22:04  #5 
1100100010010_{2} Posts 
Thanks.

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