I had earlier remarked the usual sort of remark, that the estimated probability that some DNA sequence would represent "viable life" would be ultra-extremely small if you think about the length of the DNA in shortest possible free living life form today, etc. Like, 2^(-30000). Allan Wechsler than responded he thought there might have been (what I might call) "semi life" and for this sort of vague reason the chance of abiogenesis could be a lot higher than in any such naive calculation. It now occurs to me we can construct a cheap model of such a thing and solve it, thus "proving" the chance of abiogenesis is really quite large, completely the opposite of the previous kind of naive calculation. (Probably this new calculation also is naive, but anyhow, here it comes.) ---------------- Suppose there are N different chemicals in a puddle. Among them, if two are chosen randomly (call them A & B), assume C is the (small) chance that A catalyses the formation of B. QUESTION: In this model, what is the chance F(N) that our set of chemicals contains an "autocatalytic system" i.e. (sub)set of chemicals which in net catalyse the synthesis of every chemical in that subset? (I.e, the puddle contains "life"?) SOLUTION: Equivalent problem: In a random N-vertex directed graph with arc-existence probability C, determine F(N), the chance that there exists a subgraph S each of whose vertices has invalence>=1. There are binomial(N,K) such subgraphs with K vertices (0<=K<=N). The chance that any particular one of them works (i.e. has all invalencies>=1) is (1 - (1-C)^(K-1))^K. The expected number of them that work therefore is (and probability dependencies do *not* affect this): ExpectedNumber = binomial(N,K) * (1-(1-C)^(K-1))^K. Hence log(ExpectedNumber) = log(N)*K*log(1-(1-C)^(K-1)) / log(K) approximately when N>>K>>1. Hence if N>>K>>1/C>>1 we have ln(ExpectedNumber) = ln(N)*K*(1-C)^(K-1) / ln(K) Now choose C small, K=10/C, and N=10*K=100/C, which ought to be good enough for decent approximate validity of the above. Result: ln(ExpectedNumber) = 10*exp(-10)*ln(100/C)/ln(10/C) > 10*exp(-10) > 0.000454 > 1/2203. Finally use the fact ln(x)<x to see ExpectedNumber > 1/2203. CONCLUSION ("WARREN SMITH'S ABIOGENESIS THEOREM"): Given those assumptions, among 2203 puddles we expect to find at least one that is alive! (And among, say, 10000 puddles, it is virtually certain one will be alive.) Note, this theorem does not care how tiny C>0 is, provided the puddles each contain at least N=100/C chemicals. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)