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Notices 02-13-2020, 12:03 PM #1 business_kid LQ Guru   Registered: Jan 2006 Location: Ireland Distribution: Slackware & Android Posts: 11,028 Rep:         [SOLVED] Statistics - revisited I got a statistics formula here before (Thanks guys), but now I believe what I have is/was wrong, so I have to ask again. I'm trying to calculate an estimate of the statistical probability of a particular gene being inherited for the number of generations necessary for evolution. Let's say 2500 generations for the purposes of establishing a formula. Now genetic inheritance has a probability of [P=1 in 2] or {P=0.5} per mutation, per generation. So the probability of holding 1 mutation in the first generation for 2500 generations is [P=1 in 2^2499] or {P=0.5^-2499}. OK. Here's where it gets messy: The probability of holding 2 mutations per generation is 1 in 4; of 3 mutations, 1 in 8, etc. So if we write the progression, allowing 1 mutation per 10 years, we get [P = 1 in (2)^2499+(2²)^2489+(2³)^2479+≈(2^250)^9] =1.88e752 + 3.37e1498 + 5.76e2338 + ≈2.08e677 We can simplify a little; These are brought to Scientific notation, which has only 2 decimal places, so all the minor terms are lost unless they affect the first 2 decimal places. So I would like to know how to sum that expression, short of doing it longhand. It has a geometrically progressive element to the power of an negative arithmetically progressive element  . As no evolutionist I have asked can give me any numbers, estimates, or guesses as to how many relevant mutations over how many generations/years a change of species takes, I calculate several possibilities. In the same article, I have a scenario of 1,000 mutations over 20,000 generations, and I certainly don't want to do that out longhand! Presuming that can't be done I would like to find the largest term. With additive probabilties, anything whose exponent is 2 or more smaller gets corrected out in Scientific notation, so only the near maximum terms matter. EDIT: I don't need the sum at all - I need the maximum evaluated term. Because term 2, etc includes the term 1 probability. A more correct form would be P= 1 in (2^10)+(4^10)+(8^10)+(2^250)^9. Summing that would get it. The 1st formula is right for getting the maximum term Last edited by business_kid; 02-14-2020 at 06:43 AM. Reason: solved 02-14-2020, 06:42 AM #2 business_kid LQ Guru   Registered: Jan 2006 Location: Ireland Distribution: Slackware & Android Posts: 11,028 Original Poster Rep:         [SOLVED] Statistics - revisited I have this worked out. Term n of my first equation serves as an (slightly low) approximation. The thing I asked for the sum of is actually hopelessly low. Geometric Progressions: Term n=a*(r^n-1). a=2; r=2. The Geometric bit is 2,4,8,16 etc. Arithmetic Progressions: Term n=a+((n-1)d). a=2499; d=-10. The arithmetic bit is 2499, 2489, 2479, etc. The maximum evaluated term (By trial and many, many errors) is term 126: (2^125)^(2499+(1250*-10)) = 2.04e46,998 or {4.92e-46999}

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