Philip said:
BibleWheel said:
The math is elementary. Consider a list of 39 objects. How many ways is there to divide them into three groups? The answer is easy to calculate. We need to just drop two "dividers" ^ in the list below.
1 2 3 4 ^ 5 6 7 8 ... 35 36 ^ 37 38 39
How many ways can we do this? Well, there are 38 slots to drop the first divider, and then there will be 37 slots to drop the second divider. Thus, there are 38 x 37 = 1406 ways to divide the 39 books into three groups.
Mistake #1. The grouping of books is independent of the order in which the divisions are made. Placing the first divider in slot 1 and the second divider in slot 2 yields the same result as placing the first divider in slot 2 and the second divider in slot 1. There are 703 [ C(38,2) ] combinations
Thanks for the correction! That was a silly mistake. I forgot to divide by the number of ways to arrange the dividers, something everyone learns in introductory courses on combinatorics. Like I said in a previous post, I, like all humans, make mistakes. I hope that everyone in this forum will be charitable with each other in this regard. So you are correct, the number of possible canonical structures divided into three groups is 703.
Philip said:
BibleWheel said:
But now lets consider the second level of symmetry. We do the same calculation as above, only now we drop in two more dividers.
38 x 37 x 36 x 35 = 1,771,560 ways to divide 39 books into 5 goups.
Mistake #2. (Or at least Mistake 1b.) Again, the divisions are independent of the order in which the dividers are placed. The value we want here is C(38,4), not P(38,4). There are
C(38,4) = 38! / (4! 34!) = 73,815
ways to divide 39 books into 5 groups.
Correct again, for the same reason. I like the way you noted it as "Mistake 1b" since it was a silly repetition of Mistake 1.
Philip said:
[BibleWheel said:
How many of these will show the second-level symmetry of the OT? To see this, we need to write it abstactly as follows:
Top-level n : 39-2n : n [OT has n=17 to give 17 : 5 : 17]
Second-level n-m : m : 39-2n : n-m : m [OT has n=17, m=12 to give 5 : 12 : 5 : 5 : 12 ]
Mistake #3. Your calculations assume that the first and second level of symmetry are independent events. I see no reason to assume that they are independent. Further, since you speak of a 'three-level symmetry', I doubt that you believe that they are independent.
I paramaterized the second level symmetry to impose the constraint that every solution will show both the top and second level symmetry. This does not limit the
sample space of all possible 5-group canonical structures - it limits the
solution space (see below).
Philip said:
If you want the specific symmetry you describe, not all of the 73,815 combinations we calculated above are valid. Once you've fixed the first two dividers to get your first level of symmetry, you limit the places where the third and fourth dividers can be placed.
That's why the symmetry exhibited in the Protestant OT is so rare. The constraint selects a subset (the solution space) of all possible 5-group canonical structures (the sample space).
Philip said:
We are dividing the sets of size n into sets of size n and n-m. Assuming we do not want any null sets, there are exactly n-1 ways to do this. So, while there are indeed 73,815 ways to make five sets out of 39 books, not all of those are valid members of your sample space. The sample space for your second level of symmetry is only 19 * (n-1) * (n-1)since there are 19 combinations valid for the first level of symmetry and n-1 slots in which you can place the third and fourth dividers. As you noted above, n is bounded from above by 19. So, given that the first level of symmetry is present, the cardnality of your sample space for the second level is at most 19*18*18 = 6,156. That is a far cry from the nearly 2 million you claimed above.
It seems like you are confusing the
sample space with the
solution space.
The
sample space is the set of all possible 5-group canonical structures, which we agree has 73,815 elements.
The
solution space is the set of all possible 5-group canonical structures that satisfy the following constraints:
n-m : m : 39-2n : n-m : m with 1 < n < 20 and 0 < m < n
These constraints select all 5-group canonical structures that exhibit both the top and the second levels of symmetry
without the constraint that the first division on the second level is the same size as the second division on the first level (this will be discussed below).
Philip said:
Given that the first level of symmetry exists, the odds that this second level also exists are, at worst, a mere 6/6,156.
This is incorrect. The error is that you confused the sample space with the solution space.
Now we need only to count the number of solutions that satisfy the constraints listed above. This is easy. For each n, there are exactly n-1 possibilities for m (as you correctly noted above), and n runs from 2 to 19, so the answer is .....
Sum[n=2 to 19](n-1) = 171
This number is larger than the six solutions found in the previous post because I did not apply the constraint that links the second level symmetry to the first level. This constraint demands that n-m = 39-2n (discussed below).
Therefore, the correct probablitiy (assuming I haven't made any more silly mistakes) is given by:
171/73,815 x 100% =
0.23%
That's about one chance in 431.
This is the chance that a 5-group canon would exhibit the symmetry of the Protestant OT
without the constraint that the first group of the second level is the same size as the middle group of the first level. This constaint is expressed by writing n-m=39-2n. Solving for m we have m = 3n-39, which implies 13 < n since 0 < m. Substituting this in the formula above yields:
39-2n : 3n-39 : 39-2n : 39-2n : 3n-39 with 13 < n < 20
Obviously, there are exactly six solutions, and we have the final result:
6/73,815 x 100% = 0.0081%
Alternately, we can write this as
one chance in 12,302.5
And that is only for the second level symmetry. What if we add the third? As far as I can tell, the only error in my previous calculation of this number was my use of P(38,6) rather than C(38,6) = 2,760,681 for the sample space. So the answer is:
57/2,760,681 x 100% = 0.0021%
That's
one chance in 48,433
Now lets consider what this means. All the math was really unnecessary. It should have been intuitively obvious to you, with your knowledge of combinatorics, that a three-level symmetric structure was
much less likely than a non-symmetric structure. The actual number itself is not particularly significant for the argument.
My original point stands. The three-level symmetric stucture of the Protestant OT is very rare indeed. It also is unique amongst all forms of the OT canon ever produced. Lets look at the Catholic OT as an example. It has 46 books:
5 Books of the Torah
14 Books of History (Joshua-Nehemiah, Tobias, Judith, Esther)
7 Books of Wisdom
6 Books of the Major Prophets
12 Minor Prophets
2 Books of Maccabees (the topic of this thread!)
First note that this canonical structure doesn't even display a consistent pattern of categorization. Maccabees is historical, and so is out of place with the rest of the historical books. As similar problem is seen in the Tanakh, where
Chronicles is placed
anachronistically at the end
after Ezra/Nehemiah!
But lets try to lay out the Catholic canon to see what we get:
Top-level 19 History : 7 Wisdom : 18 Prophets : 2 History
This exhibits no symmetry and a broken categorical order.
Second-level 5 Torah : 14 History : 7 Wisdom : 6 Maj Proph : 12 Min Proph : 2 History
Look at that. Each division has its own distinct number of books. There is no symmetry whatsoever. Exactly what we would expect from a chance distribution.
Third level - why bother? Its not there!
I assert that there is no other OT canon on the planet that exhibits any symmetry like that of the Protestant OT. I don't need to go through every canon for you. If you want to challenge my assertion, all you need to do is go find one that is comparable in any way to the
divine beauty of the Canon God gave us in the Protestant OT.
I really want to thank you for reviewing this with me. Iron sharpens iron!
Richard