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Old 2003-12-04, 02:02   #1
optim
 
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Default probability of finding a Mersenne prime

The Status command of the Prime95 client outputs: "The chance that the exponent you are testing will yield a Mersenne prime is about 1 in 278494".

How is this probability (1 in 278494) calculated?
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Old 2003-12-04, 04:04   #2
GP2
 
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Some of the math is discussed here:

http://www.utm.edu/research/primes/m...heuristic.html

I'm not sure if this is the actual calculation that Prime95 uses.
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Old 2003-12-06, 19:03   #3
nfortino
 
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Quote:
Originally posted by GP2
Some of the math is discussed here:

http://www.utm.edu/research/primes/m...heuristic.html

I'm not sure if this is the actual calculation that Prime95 uses.
I don't believe it is. From http://www.mersenne.org/math.htm

"What are the chances that the Lucas-Lehmer test will find a new Mersenne prime number? A simple approach is to repeatedly apply the observation that the chance of finding a factor between 2X and 2X+1 is about 1/x. For example, you are testing 210000139-1 for which trial factoring has proved there are no factors less than 264. The chance that it is prime is the chance of no 65-bit factor * chance of no 66 bit factor * ... * chance of no 5000070 bit factor. That is:

64 65 5000069
-- * -- * ... * -------
65 66 5000070

This simplifies to 64 / 5000070 or 1 in 78126. This simple approach isn't quite right. It would give a formula of how_far_factored divided by (exponent divided by 2). However, more rigorous work has shown the formula to be (how_far_factored-1) / (exponent times Euler's constant (0.577...)). In this case, 1 in 91623."

Doing the calculation on both my computers, I come up with a rather close number using the formula with Euler's constant.
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