From abdulrahman@worldnet.att.net (AbdulraHman Lomax) Newsgroups: soc.religion.islam Subject: Re: Anyone tried this as yet ? Date: Sat Aug 16 18:14:53 EDT 1997 Message-Id: 5t58ot$jkv@usenet.srv.cis.pitt.edu as-salamu 'alaykum. Fred Karmallywrote: >Basem Mesbah Temraz wrote: >> I believe that the biggest problem seems to be that the method of >> counting is not consistent. i.e. the rules of counting seem to >> change from word to word so that it can be produced in multiples >> of 19. Br Temraz was accurate in his description. We do not ordinarily stop to think about how many different ways things can be counted and analysed. Let's see what Fred does with this: >Does the following seem to be inconsistent ? >64 380 >48 276 >300 53 >44 324 >16 150 >31 200 >36 225 >These are the frequencies of the H and M in the 7 chapters entitled H.M. Now, Fred has not been precise here. First of all, only one chapter is "entitled H.M." But there are seven chapters with "H.M." in the first verse. One of these chapters has an additional three initials in the second verse. Or is it the third verse? Depends on whether or not one counts the bismillah, doesn't it? Is the bismillah counted in the above counts? As is typical with Khalifite reports, we are not told. Yes, they are counted. Also, Fred gave the wrong values in the third line, which should be fairly obvious: he reversed the Ha and Mim numbers. Now, Fred proceeds to do his number magic. Actually, he did not invent this trick, others came before him. >Observe the following , >1) sum of rows 1,2 and 3 = 59 x 19 > sum of digits in rows 1,2 and 3 = 59 >2) sum of rows 4,5,6,7 54 x 19 > sum of digits of rows 4,5,6,7 54 >3) sum of rows of rows 2,3,4 55 x 19 > sum of digits of rows 55 >4) sum of digits of rows 1,5,6,7 58 x 19 > sum of digits of rows 1,5,6,7 58 There are three parts to this trick. First, the methods of counting were varied until a method was found that produced a total count for H and M in the seven suras that was a multiple of 19. It might seem that the count of the letters would just be a "fact," and that it could not be manipulated. But one who actually does research in these kinds of things soon finds that there are many choices to be made. First of all, is the bismillah included or not? Dr. Khalifa, in his counts of word frequencies, did not include the bismillahs after the first, except for the one occurrence in the letter of Sulayman. But when he counted initial letters in suras, he counted the initial bismillah, at least he did in his final counts; he went back and forth over the years on this point. No other initial letters are counted and combined in exactly the same way as these letters in the H.M. series. With each set of initial letters, almost always, a different method of counting is used. To give another example of a choice, Sura 42, beginning with "H.M.", also has "Ayn.S.Q.", so one could include or exclude it on that basis. With a few choices, it becomes fairly easy to find a combination that will be divisible by 19. (NOTE: I'll call this "ZMN", for "zero modulo nineteen," which is mathspeak for this phenomenon.) To quantify this, the probability that a random number will be ZMN is about 0.053. Now, if I have a collection of letters to count, and there is more than one possible way to count them, the probability that there exists a way that produces a ZMN total will increase. Here is the probability that at least one way will be ZMN for a given number of ways to count: 1. 0.0526 2. 0.1025 3. 0.1497 4. 0.1945 5. 0.2369 6. 0.2770 Now, suppose that there are independent choices that are made in counting. Suppose I were counting people. I could count the men, or the people over a certain age, or those with blue eyes. These choices are independent, so they can be combined to produce different totals. So two independent binary choices gives *four* ways to count. Anyway, the first part of the trick is that the total for *all* H. and M., including the bismillahs, in the suras 40 through 46, which is all the suras initialled with Haa Miym, is ZMN. It is a fact that is somewhat interesting, though certainly not remarkably unusual. Fred did not mention this,but it was the first of these "facts" to be discovered. By the way, I am relying on Khalifa's final totals here; he published many erroneous totals (each time finding some miracle in his incorrect numbers) before finally settling on these totals, which I suspect are correct. Now, suppose we have seven numbers like this: 7, 1, 11, 7, 14, 3, 14. This is merely the set of remainders after the H.M. totals for each sura are divided by 19. These add up to 57, which is three times 19, and that the total would be ZMN, of course, follows from the fact that the original numbers have a ZMN total. This brings us to the second part of the trick. If we have seven numbers whose total is ZMN, what is the probability of finding subsets which are ZMN? How many subsets are there? The smallest subsets have only one member, and there are obviously 7 of these. Then, if the subset has two members, there are 21, which is the number of ways that one could combine two objects out of seven. (This is how many two-letter combinations one could make out of the letters ABCDEFG, if one really wants to verify my result. Note that AB would be the same combination as BA....) number of members number of subsets 1 7 2 21 3 35 4 35 5 21 6 7 The symnetry of this list is normal; it is called, I think, the binomial distribution. Anyway, for every set of three or fewer members, there is a corresponding set with four or more members, that is, the *other* members of the original set of seven. And if one of these sets is ZMN, so is the other. So what looks like four "facts" of 19-divisibility is actually just three. There is the original, unstated fact that the sum of rows 1 through 7 is ZMN, and then we have rows 1,2,3 giving a ZMN total (from which it necessarily follows that rows 4,5,6 and 7 with a ZMN total), and then we have rows 2,3,4 with a ZMN total (and then 1,5,6,7 follows). How likely is it that two subsets of seven rows of random numbers would add up to a ZMN total, given that all the numbers give a ZMN total? Pretty likely. We can see that there are 63 different pairs of subsets from the above table. If one takes 63 random numbers, what is the probability that two of them are divisible by 19? There are ways to calculate this, but I am not going do it here; I will only note that I did a computer simulation some time ago on this problem and found that it was more common to have *three* pairs of ZMN subsets than only two. Now, we come to the third part. If a set of numbers add up to a multiple of 19, what is the probability that the sum of digits for those numbers add up to the total of the set divided by 19? Ah, this poor old brain; it has done good service for many years, but it is getting worn in parts. Not to mention somewhat lazy. So I will merely assert that my experience with number congruencies, which began nearly forty years ago, leads me to suspect that it is fairly high, if the numbers are within a certain range; and if there is a table of such numbers, all more or less in the same range, as there is above, if one set displays this phenomenon, the others are likely to do it as well. >Did you know that to come up with a nonzero average solution of 150, >you would have to do 14 nested iterations from 1 to 150 and perform >84 computations each time (for 2 digit no, 3 digit would be even more). I don't even know what a "nonzero average solution of 150" is, much less how would come up with it. I suspect that Fred is quoting someone else who perhaps had a little more knowledge of what he was talking about. But I have never seen a Khalifite write on the subject of the probabilities of the phenomena described who actually made a correct analysis, and some of the published material, such as the little booklet on the Bismillah by Abdullah Arik, is seriously defective, vastly overstating the rarity of the phenomena (by treating the experimental conditions as if they were random. What is the probability that my social security number begins with a 6? Well, one might say that since there are 10 different digits, 6 is one out of ten, and the number beginning the social security number is one out of ten, so the probability that they are the same would be 1 out of 10 x 10, or 1 out of 100. Ark repeated that error numerous times, so 1 out of 19 became something like one out of 19 million, as I recall.) >This equals 26 x 10**31, i.e. a computer doing 1 trillion calculations >per second would take about over 1 trillion years to do this. This does not appear to follow from what was said previously. >Never mind that these are frequencies of letters within words that not >only make sentences but are of the highest linguistic quality. A Freudian slip, perhaps. After all this discussion of the supposedly miraculous nature of these letter counts abstracted from the Qur'an, Fred tells us to "never mind", ultimately, the *sentences* of the book. Is it the letters or the words which are "of the highest linguistic quality"? Or is it the *sentences* or even something on a more integrated level? Certainly my letters are of the highest quality found in the English language, all 26 of them; are we to assert that the letters of the Qur'an are also the highest quality letters in Arabic? I am sure that they are. Now, just what do we do with these letters? One can understand speech without knowing anything about letters. And one can take letters and put them together in different patterns and say the opposite of what was originally said with them. Some Satanists might read a religious text backwards, and surely it is still composed of the same letters when it is backwards.... >Praise God Indeed. AbdulraHman Lomax abdulrahman@worldnet.att.net P.O. Box 10316 San Rafael, CA 94912

Overview on numerical features in different scriptures

Answering Islam Home Page