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In article <4naaui\$dr0@usenet.srv.cis.pitt.edu>, Brendan McKay  writes in response to:

> MORE MATHEMATICS:
>
> "Al-Wahid" "The One" referring directly to one of God's most
> beautiful names is mentioned in 12:39, 13:16, 14:48, 38:65,
> 39:4 and 40:16
> If we add all these numbers of the Verses and Suras we get :
> 12+39+13+16+14+48+38+65+39+4+40+16=344
>
> This number, 344 is a composite number whose index is 275
> (in other words 344 is the 275th composite number)
> 344+275 = 619   Here again God is posting his 6 and 19 in
> relation to His Oneness in a more sophisticated system.

This is truly amazing.  All that work and you STILL could not
find a multiple of 19.  I think this deserves this week's
Futility Prize, congratulations!

Let me take this opportunity to inject some probability facts.
A favourite device of the 19ers is to add things together to
make 19 when the items themselves don't work.  For example, if
the sura number and the verse number have no 19, maybe their
sum does.  Khalifa does this quite a lot, even adding three
things if necessary, for example verse number + number of words
+ number of letters.  He still counts the result as having a
probability of 1/19, but of course that is wrong.

Suppose you have N numbers and you want to find a multiple of 19 by
adding some of them together.  What are the chances of success?
This calculation is a little messy, but I have done it up to N=5:

N    probabilty of finding 19

1          5.3%
2         15.2%
3         32.8%
4         59.2%
5         86.4%

So, for example, if you have the sura number, verse number, number
of words, number of letters, and gematrical value of a particular
verse, there is an 86.4% chance you can get a multiple of 19 by
adding some of them together.  You have to be very unlucky to fail!

-------------------------------

Where do these probabilities come from? Consider for N=2:
A = the event that the first number is a multiple of 19,
B = the event that the second number is a multiple of 19,
C = "A + B" = the sum of both numbers is a multiple of 19.

If any two (of the three) are multiples of 19, so is the other.

Write P(X) for Prob(X is a multiple of 19) etc.

P(A or B or C) = P(A) + P(B) + P(C)
- P(A and B) - P(A and C) - P(B and C)
+ P(A and B and C)

Each of P(A), P(B), P(C) are 1/19, and each of the others are 1/19^2.
So the answer is 3/19 - 3/19^2 + 1/19^2 = 3/19 - 2/19^2.

Another easier method: Find the probability that none are multiples
of 19. For A, choice is 18/19.  Given A, the choice for B is 17/19:
one 1/19 is lost to ensure B is not a multiple of 19, and another 1/19
is lost to ensure A+B is not a multiple of 19.  These two cases do
not overlap because at this point A is not a multiple of 19.

I did it like that for N=3 but already it is hard to get correct.
After that I wrote a program just to count all the cases.

Brendan
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