Non-Random Equidistant Letter Sequence Extensions in Ezekiel Continued By R. Edwin Sherman, FCAS, MAAA, and Nathan Jacobi, Ph.D.

 Appendix B—A Markov Chain ELS Extension Model This addendum provides a derivation of formula (1), which is used to determine the expected number of extended ELSs (with varying numbers of extensions) to emerge from a search process beginning with n initial ELSs. We shall denote initial ELSs by the symbol I. After a Hebrew expert examines the string of letters resulting from taking every j-th letter before and after the initial ELS with a skip of j letters, one of two results will occur in each instance that there is an opportunity to find an extension of the initial ELS. If no extension is found, this will be denoted by N and if one is found, by E. The following chart presents the complete range of outcomes from this process up through the location of four extensions to an initial ELS. From this chart, the formula for determining the expected number of ELSs with a given number of extensions is evident. Each outcome is represented first by a combination of the letters I, E and N, indicating the order in which these events appeared in that letter string. In the middle section of the table is a row of formulae for the expected number of ELSs for each type of outcome in that column. In each formula, n is the total number of initial ELSs, d is the probability of finding a grammatically correct Hebrew extension of the preceding ELS, and (1-d) is the probability of failing to find an extension. It was assumed that d is independent of the number of extensions that have already been discovered—even though consideration of the different factors affecting d as the number of extensions increases suggests that d most likely declines as the ELS becomes longer.

 The beginning of the search process is represented by the leftmost column of the table, where there are n instances where I, an initial ELS, appears. Each column represents the range of outcomes that would result in the indicated number of final extensions. The only way that zero extensions will be found is the situation where no extension is found either before or after the initial ELS (denoted by NIN). The probability of this occurring for any given initial ELS is (1-d)2. So the expected number of final ELSs that have no extension is n(1-d)2. There are two ways that a final ELS can have one and only one extension: NIEN and NEIN. The expected number of final ELSs that are of the NIEN type is nd(1-d)2, as is the case for the NEIN type. So the total expected number of final ELSs with exactly one extension is 2nd(1-d)2, as shown at the bottom of the "One" column. There are three ways that a final ELS can have exactly two extensions: NIEEN, NEIEN and NEEIN. For any given column (representing k extensions), I can appear in the 1st, 2nd… or (k+1)st position after the first N. Thus the total expected number of final ELSs with exactly k extensions will be (1)                                                     (k+1)ndk(1-d)2. The above model is an example of a Markov chain, since the outcome of any trial depends at most on the outcome of the immediately preceding trial and not upon any other previous outcome, and the probability of each state is clearly defined. When the state N occurs, it is an "absorbing state." As a check on formula (1), it should be possible to show that the sum of (1) for k ranging from 0 to infinity is n, the total number of initial ELSs. That derivation is as follows: Since n and (1-d)2 appear in each term, we can factor them out of an expression for the total number of final ELSs, to get n(1-d)2 [Ó (k+1)dk], where k ranges from 0 to infinity. If we multiply each term of this power series by d in both the numerator and denominator, we obtain {[n(1-d)2]/d} [Ó (k+1)(d{k+1}]. If we shift the value of (k+1) by one for each term, so that the series is summed from k = 1 to infinity, rather than from k = 0 to infinity, it becomes (2)                                                 {[n(1-d)2]/d} [Ó kdk]. According to formula 40 on page 8 of Summation of Series, collected by L.B.W. Jolley: Ó nxn = x/(1-x)2, where x < 1, and the series is summed from n=1 to infinity. If we substitute k for n and d for x in this formula, we have Ó k(dk) = d/(1-d)2. By substituting the expression on the right for the power series expression in (2) above, we get {[n(1-d)2]/d} [d/(1-d)2] = n.

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Code Skeptics' Arguments Trashed

Ever since the first Bible codes were announced, skeptics have been saying, "Oh, well, you can also find codes like that in books like War and Peace and Moby Dick."

We took the time to examine this notion and the best example of it that the skeptics have been able to come up with. The results of our research have completely blown away their theory.