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 Karmally wrote:
>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
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