## Re: [GTh] The Literary Unity of Thomas

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• Mike, Basically what I did was cast a reasonably large net, thus I don t assume perfect parallels (although they appear to exist in this case), rather I wanted
Message 1 of 11 , Dec 3, 2009
Mike,

Basically what I did was cast a reasonably large net, thus I don't assume perfect parallels (although they appear to exist in this case), rather I wanted to see what the probability was that two thematically paralleled logia would appear in the first eight logia AND the last eight. Thus my math allows each logion to be positioned anywhere inside the first and last eight. I did so, like I said, to cast a broader net while still showing statistical significance. This equation I used is given below with the first 8/114 being the likelihood of one of the sayings landing in the first 8, 8/106 is the probability the other lands in the remaining 106. If we crunch the math like you suggest, looking at the probability that both fall exactly 8 sayings from the beginning or end, the chance that there were randomly put there falls to something like 2x10-4, i.e. almost infinitesimally small.

(8/114)*(8/106)*2 = 128/12084 = 0.0105925

ian

________________________________

> The probability that these almost identical sayings would randomly be
> placed in the first eight, and last eight sayings respectively is
> 0.0105925
> (or 128/12084) ...

Could you explain this calculation, and also the amended calculation
given the corrected position of L107?

Mike

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• ... Hi Ian, In the first place, I m not sure why you say if , since the two sayings are in fact eighth from the beginning and end, respectively. Secondly, the
Message 2 of 11 , Dec 4, 2009
> If we crunch the math like you suggest, looking at the probability that
> both [L8 & L107] fall exactly 8 sayings from the beginning or end, the
> chance that there were randomly put there falls to something like 2x10-4,
> i.e. almost infinitesimally small.

Hi Ian,

In the first place, I'm not sure why you say 'if', since the two sayings
are in fact eighth from the beginning and end, respectively. Secondly,
the math doesn't seem right to me. Suppose, for example, that ALL
of the sayings were paired up, i.e., 57 pairs of sayings, but in random
order. Wouldn't it be more likely that at least one of those pairs (say Ln
and Lm) would be such that Ln would be the same distance from the
beginning as Lm was from the end, than if there were only a few such
pairs? But if that's so, then it seems that the calculation of random
probability should yield a different result depending on the frequency
of such pairs. But I think the algorithm you're using doesn't do that;
I think it assumes just one such pair.

Cheers,
Mike
• Hi again Ian, After sending off my previous note, an even more radical criticism of the random-probability algorithm occurred to me. Let s suppose that L7 and
Message 3 of 11 , Dec 4, 2009
Hi again Ian,

After sending off my previous note, an even more radical
criticism of the random-probability algorithm occurred to me.
Let's suppose that L7 and L108 are the only such pair in 114
sayings and that the sayings are randomly rearranged. Now
wherever L7 is placed, there's a choice of 113 positions for
L108, one of which will be the same distance from one end
as L7 is from the other. That would seem to indicate that there's
a 1 in 113 chance that L7 and L108 will be equidistant from
beginning and end. That's the probability if there's only one such
pair. If there's more (and there are), the probability of any one
of them being so arranged is less than 1 in 113. Or so I figure.
Do you see anything wrong with that?

Mike
• Mike, I think it is you now who are playing on the conservative side. For in your math you presume an intentional placement of the first logion which then
Message 4 of 11 , Dec 4, 2009
Mike,

I think it is you now who are playing on the conservative side. For in your math you presume an intentional placement of the first logion which then leaves 113 possible placements for the second. But this is not the way random distribution works because it would present BOTH logion 7 and 108 as being randomly distributed. Thus the probability of them being in the 8th and 8th last (or first or last, or whatever two positions) would actually be (1/114)*(1/113)*2 which is, as I indicated in a previous note, is 1.5x10-4, or .00015/1.

So here I will try to clarify what I did, and what you call for. What I did was look at the probability that one falls within the first eight, and one falls in the last. I did this because I see many parallels in this body of Thomas (3//113, 5//108, 6//104, and 8//107) but only one is actually a perfect mirror. Thus while I think parallels are present, I don't think there is a chiasm. That is why I didn't bother looking at the probability of two sayings directly mirroring each other positionally (which you call for, and which is certainly the case with 8 and 107).

Does this make sense. To put my math into an example. If I shuffled114 cards, 112 of which were blue, and 2 of which (logia 7 and 108) were red, the probability that one red card would fall in the first eight while the other fell into the last eight is 128/12084.

Regarding your appeal to multiple pairs, I'm not quite sure what you're referring to. Could you elaborate?

ian

________________________________
From: Michael Grondin <mwgrondin@...>
To: gthomas@yahoogroups.com
Sent: Fri, December 4, 2009 12:01:09 PM
Subject: Re: [GTh] The Literary Unity of Thomas

Hi again Ian,

After sending off my previous note, an even more radical
criticism of the random-probability algorithm occurred to me.
Let's suppose that L7 and L108 are the only such pair in 114
sayings and that the sayings are randomly rearranged. Now
wherever L7 is placed, there's a choice of 113 positions for
L108, one of which will be the same distance from one end
as L7 is from the other. That would seem to indicate that there's
a 1 in 113 chance that L7 and L108 will be equidistant from
beginning and end. That's the probability if there's only one such
pair. If there's more (and there are), the probability of any one
of them being so arranged is less than 1 in 113. Or so I figure.
Do you see anything wrong with that?

Mike

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• Ian, I wanted to let you know that I m mulling over the meaning and implications of our two methods of calculation. I started writing a response, but wasn t
Message 5 of 11 , Dec 5, 2009
Ian,

I wanted to let you know that I'm mulling over the meaning and
implications of our two methods of calculation. I started writing
a response, but wasn't satisfied with it, so this will have to do for
the time being. Where I do agree with you is that your formula
captures the probability of two given sayings occurring at 1+n and
114-n, for any given n. Still, I can't shake the feeling that my
calculation has something right about it. What I think it is (though
I can't quite get it straight yet) is that my formula may relate to
_some n_, versus _any n_. (Ex: the probability of any one of the
five horses in a 5-horse race winning the race may be 1/5, but the
probability that _some_ horse will win is 100%.)

As you rightly say, your formula relates to any n at all, but what I would
add is that it's only one n. Each and every n has the same probability,
but that doesn't seem to tell us what the probability is that there's
_some n_. Intuitively, I think that's what we're looking for, and the
probability of that should be greater than the probability of any single n.
That's where I think my formula comes into play, and I sense that it's
really that probability that's needed to make an accurate estimate of
intentionality.

Cheers,
Mike
• Hi again Ian, It s early morning here (about 1:30 am and probably an hour later before I finish this note), but I don t get much chance to write during the
Message 6 of 11 , Dec 5, 2009
Hi again Ian,

It's early morning here (about 1:30 am and probably an hour
later before I finish this note), but I don't get much chance to
write during the day, and I wanted to pass along a couple of
thoughts while they're still fresh in my mind, even though I'm
not in tip-top mental shape at this late hour. These thoughts
relate to the question you posed some time back, viz. whether
the placement of L8 and L107 was intentional. I believe that
it was, but my reason for believing so isn't based on the
probability calculations we've been discussing. What I want to
do in this note is two things: (1) to show that probability calcula-
tions don't provide strong enough support for intentionality, and
(2) to present some new considerations that I believe do provide
the needed support.

First, as to the calculations:

C1: (1/114)*(1/113)*2
C2: 1/113 [as original, but now amended to (1/113)*2, or 2/113]

C1 gives us the probability of any two sayings occuring at, say, 1+n
and 114-n (i.e., equidistant from the first and last sayings, where n
is possibly 0). So it tells us in particular how likely it is that any two
sayings might randomly appear both 8th from the top and 8th from
the bottom (i.e., n=7). But the same formula also tells us how likely
it is that those two sayings might appear 7th from the top and bottom,
or 13th, or 57th. Unless there is something special about being 8th,
(and just to let you know, I will argue below that there is), it's no more
meaningful that the pair should be 8th than that they should be x-th.
But in that case, the probability that we're interested in isn't C1, but
C2. For C1 reduces to C2 (by multiplying C1 by 114) when what we're
interested in is the probability that there's _some_ such random
positional relationship between the two sayings, whether it's 8th, 7th,
13th, or 57th.

What it comes down to, I believe, is that the odds of L8 and L107
(or indeed, any two sayings) being equidistant from top and bottom
in a random distribution are 2 in 113, or about 1 in 57. Not great odds,
to be sure, but a far cry from astronomical. It was unlikely to have
happened, but there's an almost 2% chance that it did.

Now here's the good news: the case doesn't need to rest on
probability, because there are _other_ reasons to suppose
that the placement of L8 and L107 was intentional. In fact, the 8th
position _does_ appear to be special. Not just because in 1+n and
114-n, n=7, which can be shown to be a number of special interest
to GosThom in both saying and construction, but because if the
seven sayings which precede L8, and the seven sayings which
follow L107, are removed, what is left is 100 sayings. That result
is suggestive in itself, but it's startling that L107 - which would be the
last of the 100 sayings (8-107) - is specifically about a man who
had 100 sheep. To me, this is positive textual confirmation that
we're on the right track.

Mike
• I know this conversation is a few days old at this point, but because I m a math nerd, I feel like I have to chime in. The calculation for the probability that
Message 7 of 11 , Dec 7, 2009
I know this conversation is a few days old at this point, but because I'm a math nerd, I feel like I have to chime in. The calculation for the probability that L8 and L107 would be paired together in any perfect mirror out of all the sayings is:

57 * 112! 57 1
P = -------- = --------- = --- = 0.44%
114! 114 * 113 226

Explanation below:

Think of a row with four chairs. If I have four people, and I want to figure out the total number of ways to arrange those four people, I would multiply

4 * 3 * 2 * 1 = 4! = 24

because I can put any of the four people in the first chair, then I have three remaining people to choose from in the second chair, then two people for the third chair, and finally my last person in the fourth chair. In the case of Thomas, it's like I have 114 chairs for 114 sayings. So the denominator of the probability formula above:

114!

...gives us the total possible arrangements of the sayings.

To get the numerator (arrangements wherein L7 and L108 form a perfect mirror), I have to think through how often they could be in a perfect mirror at all. And there are 57 ways for that to happen (1 and 114, 2 and 113, 3 and 112, etc.). But in each of those 57 perfect mirrors, I still have 112 sayings to place. And since there are no qualifications for how to arrange those other 112 sayings, I can perform the same calculation as above. Except now I have 112 spots for 112 sayings, yielding:

112!

So, for each of those 57 times that L7 and L108 fall into a perfect mirror of each other, there are 112! ways to arrange the other sayings. Meaning our numerator (the number of random distributions that put L7 and L108 in a perfect mirror) is:

57 * 112!

Put the numerator and denominator together and you get the probability above. Now, some assumptions and caveats.

First of all, we have to assume that we want L8 and L107 in particular; that is, that there are no other logia that could be thought of as parallel. For every logia that might also pair up with L7 and L108 (e.g., L76), the probability that a parallel pair will be perfect mirrors of each other goes up.

Second of all, the above calculation assumes that the mirror can occur anywhere in the 114 sayings (1 and 114, 5 and 110, 50 and 65, etc.). If you want the probably that it occurs in a particular area, you would multiply by that area and divide by the total (57 possible pairings). For example, if I think the first eight sayings are particularly significant, I would take my 1/226 probability for a perfect mirror occurring anywhere in the 114 sayings and multiply by 8/57 (the area of eight pairs I'm interested in divided by the total of 57 pairs). That yields a probability of:

1 8
P = --- * -- = 0.062% (slightly more than 6 out of 10,000 times)
226 57

Third of all, look at how we have to talk about the distribution of the sayings to make this all work in the first place. The only thing that the above probabilities demonstrate is what chance there would be in a random distribution of 114 sayings for L7 and L108 to form a perfect mirror. The calculation assumes complete independence of the 114 logia, as if they were all separated out onto different slips of parchment then shuffled together at random, then recopied into the Coptic Thomas we have today. And, of course, that doesn't remotely resemble any process we know of for the production of ancient texts. Moreover, as with the versification of the Bible, Thomas was not originally written with our modern 114 separations in mind. So any numerically patterns we find within those modern 114 distinctions needs to be tempered carefully by the recognition of that fact.

Fourth of all, as Michael aptly observed earlier, probability does not imply intentionality. Particularly when we already have the Thomas text. It would be like walking up to someone, asking them their birthday and telling them yours, and then saying "Wow! What are the chances that you and I, with our birthdays, would walk up to each other today? Only .00075%!! Why, this can't be a coincidence!" Except that you've already walked up to each other. And you both have to have birthdays. So no matter who you would've walked up to, there'd be a .00075% chance of having those two particular birthdays. In the same way, we already have the text of the Gospel of Thomas. Probability won't tell us very much about that text.

[Tim]
• ... Hi Tim, and thanks for your thoughtful message. As it happens, I was intending to write in tonight to correct a little something, and your note provides a
Message 8 of 11 , Dec 7, 2009
[Tim's calcs]:
> 57 * 112! 57 1
> P = -------- = --------- = --- = 0.44%
> 114! 114 * 113 226

Hi Tim, and thanks for your thoughtful message. As it happens, I was
intending to write in tonight to correct a little something, and your note
provides a perfect excuse. The little something I was going to correct
was the '*2' that appears in Ian's calcs and which I tentatively adopted
in my last note. I now think that that's wrong, and the reason I think so
is related to your calcs also.

OK, here goes: I believe that the '*2' was put into Ian's calc to account
for the possibility that the given two sayings might appear in two
different orders at the same positions. Ln, for example, might occur
in the 5th position and Lm in the 110th, but also Lm might be in the
5th position and Ln in the 110th. In both cases, the sayings are the
same distance from top and bottom.

However ... the 114*113 part of the calc already takes account of
this. For it matches position 5 with position 110 and then again 110
with 5. So there's no need for the '*2' and in fact that makes the calc
wrong. As far as that goes, I come back to my original 1/113 probability.
Now it remains to show that your calc contains an opposite error of the
same type, and when that's corrected, yours also comes out to 1/113.

I assume that your factorials are correct. What's in question is the
number 57. I believe it should be 114. And as you can see, if the
numerator of your second formula is 114, it reduces to 1/113.
My reasoning is this: the denominator (114*113) contains every
possibility, including order-reversals. So as above, position 5 would
be paired with 110 and then again 110 would be paired with 5. But
the number 57 _doesn't_ contain order-reversals. It counts 5 and 110
only once, not twice. Now we could go either way - i.e., we could
count order-reversals or not. But whichever way we choose to go,
we have to be consistent throughout the formula. For the above
numerator be consistent with the denominator, ISTM that it would
have to be 114. QED, I believe, but let me know what you think.

> First of all, we have to assume that we want L8 and L107 in particular;
> that is, that there are no other logia that could be thought of as
> parallel. For every logia that might also pair up with L7 and L108 (e.g.,
> L76), the probability that a parallel pair will be perfect mirrors of
> each other goes up.

I agree, and would just add that even if there were no three-somes,
the number of pairs like L8 and L107 would affect the picture. The
more mirror-doublets (or whatever you want to call them) there are,
the more likely that one of them will occur equidistant from top and
bottom. I didn't want to get into this, however, since we haven't yet
agreed on even the basic calcs for one pair.

> The calculation assumes complete independence of the 114 logia,
> as if they were all separated out onto different slips of parchment
> then shuffled together at random, then recopied into the Coptic Thomas
> we have today. And, of course, that doesn't remotely resemble any
> process we know of for the production of ancient texts.

I don't think we really know anything about how sayings-lists were
assembled, however. Besides, the calcs are _supposed_ to assume
independence, because they're a measure of randomness, not of
the way the text is necessarily assumed to have been assembled.

> Moreover, as with the versification of the Bible, Thomas was not
> originally written with our modern 114 separations in mind. So any
> numerically patterns we find within those modern 114 distinctions needs
> to be tempered carefully by the recognition of that fact.

To my mind, that's more an assumption than a fact. The assumption is
based on the fact that the text isn't numbered, and that's normally good
enough. But this is an odd text the use of which isn't clear, and I would
therefore hold open the possibility that a student-reader might have been
intended to interact with it in some way, and that that might have included
counting the sayings.

Best wishes,
Mike
• To Tim, Ian, et al: I ve been mulling over (some would say obsessing about) this probability thing for five days now, since Ian first posted. Every time I
Message 9 of 11 , Dec 8, 2009
To Tim, Ian, et al:

I've been mulling over (some would say obsessing about) this
probability thing for five days now, since Ian first posted. Every
time I write in, I believe I've finally got it right, but then thinking
about it afterward, I realize that I've bungled somewhere along
the line. In the present instance, the statements in my last note
about the import of '*2' in Ian's formula aren't right. Let me try again.

I start with the clear idea that the number of possible positions
for two sayings within a collection of 114 sayings is 113*114.
This includes positional reversals, i.e., both "n,m" and "m,n"
for any two different numbers in the range 1-114. In particular,
it includes '8,107' and '107,8'. It follows that the probability of
L8 and L107 occurring randomly exactly where they are is
1 in 113*114. If, however, we are interested in the probability
of their occuring 8th from the top and bottom, regardless of
order, then the probability is _2_ in 113*114, because not
only might they occur exactly where they do, but also L107
might have occurred in position 8 and L8 in position 107.
So Ian's formula is correct for that particular probability.

Mike G.
Mt. Clemens, MI
(Where the first snow of the season is upon us!)
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