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Receiving 1 moving line for a hexagram

peter

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I see some calmness on this forum in the last 1,5 months. Well, I think I can offer something interesting to discuss.

I found an interesting method of casting a hexagram with 1 and only 1 moving line, so it is very useful just for textual methods; you know many difficulties which occur when one gets many moving lines, there are some rules for choosing only one line from them etc. Method with 1 line is based on traditional method of casting a hexagram using yarrow-stalks, and in this article I?ll base just on this method with its odds in getting greater/lesser yin and yang lines. I read that this method was reconstructed by scientists in Nanjing some 20-30 years ago. Hope I?ll find more on it in future.

So firstly we cast a hexagram, writing down lines with numbers 6, 7, 8 or 9. Note: write statical hexagram (6 and 9 are not moving lines here). Then calculate total sum of all the numbers of lines. You?ll receive a number from 36 (all sixes) to 54 (all nines). Finally, pick up your moving line according to this table:

6th line: 41, 42, 53, 54
5th line: 40, 43, 52
4th line: 39, 44, 51
3rd line: 38, 45, 50
2nd line: 37, 46, 49
1st line: 36, 47, 48

Looks nice, doesn?t it? You always get 1 of 384 line texts along with your hexagram. And you can use it for analyzing a hexagram by Shao Yong?s method (Plum Blossom).

Now I want to play with numbers and probabilities (that?s why I insisted on traditional yarow-stalk, for coins probabilities will be different). Firstly, let?s ask ? what are probabilities to receive top moving line or say 2nd moving line? I made some calculations which I want to show here (briefly, of course). Totally, there are 4096 variants which give us those 19 sums. 4096 is 4 (amount of variants for 1 line) raised to the sixth power (by amount of lines). Yes, calculating all these variants would be boring, so I engaged combinatorical analysis into my calculations and got 84 combinations of all the 4 numbers (6, 7, 8 and 9). You see, we don?t need to take into account an order in which our numbers go in a hexagram cast, we are interested in sum only. There is only 1 variant of all sixes, or all nines etc. But if we meet 1 number of another sort, then we have 6 variants for this set (say 877777, we can cast 778777 or 777787, but their sum is a constant); if we meet 2 numbers of one sort and 4 of another, then we have 15 variants etc. If somebody is interested in these basical combinatorical things called ?combinations?, I recommend to examine some courses on combinatorics. After calculating of these amounts (they should make a sum of 4096, surely, and they made it) I also calculated probabilities of receiving of every set (6 has 1/16 chance to meet, 7 ? 5/16, 8 ? 7/16 and 9 ? 3/16). So I aligned them all by those 19 sums from 36 to 54 and got the following percentage (I had to divide on 17 millions almost ? 2 in 24th power ? that?s why some percents are so small):

36 0,000006%
37 0,00018%
38 0,00249%
39 0,02%
40 0,13%
41 0,54%
42 1,78%
43 4,55%
44 9,16%
45 14,66%
46 18,70%
47 18,98%
48 15,24%
49 9,56%
50 4,58%
51 1,62%
52 0,40%
53 0,06%
54 0,00435%

The most probable is the sum of 47 (say five 8-s and one 7). Note that for coins this chart would be symmetrical with center in 45.

So let?s sum these percents according to the first table and see the results:

6th: 2,39%
5th: 5,08%
4th: 10,80%
3rd: 19,24%
2nd: 28,26%
1st: 34,23%

Very irregular picture. The sum of 47 falls on 1st line. And 1st line is almost 15 times more probable to get than top line! Isn?t it too odd? I see two ways: 1) to reconcile to such a dispersion and take into account that when we?ve got top line moving it is a very rare event, and it can mark either very good luck or very bad luck according to its text (for every hexagram there will be its own probabilities of receiving this or that moving line though, but I believe they don?t differ much from the total percents); 2) rearrange those sums for lines so these odds will be as minimal as they can be. I want now to continue with the second way and finish my calculations. But these odds will remain, because that 47 is too huge number to cover it (totally even system would give us 16,67% for every line).

We have 2 tasks here: a) minimize oddities (sum big percents with small, and average with average), and also b) ensure that we are able to receive every 1 moving line for every hexagram. The second task is not a simple one. Let?s examine it. Every hexagram can give us 7 numbers: say ?The Well? statical gives us 8+7+7+8+7+8 = 45. Every line can change its number by 2 (8 to 6 or 7 to 9, or it won?t be ?The Well? otherwise), so if all 8-s change to 6-s we have the sum of 39, and if all 7-s change to 9-s we have 51. So ?The Well? covers these 7 sums: 39, 41, 43, 45, 47, 49 and 51 (while there are 7 cases: from all ?young? lines ? 0 old ? to all ?old? lines ? 6). Pay your attention that a hexagam can have either even (the rest from dividing by 2 is 0) or odd (the rest is 1) sums. This rule is considered in the initial system of calculating that 1 moving line. Resume: all hexagrams cover consequent 7 even or odd numbers. These multitudes intersect in the following numbers: 41, 42, 43, 44, 45, 46, 47, 48 and 49. I mean that in every set of 7 sums for one hexagram we will meet either even or odd sums from this row (42, 44, 46 & 48 or 41, 43, 45, 47 & 49). What does it give us? An important thing: these numbers must be distributed by different lines. Okay, while we have 6 lines and 9 common numbers, it is not so hard. But the rest 10 sums we must distribute in that way they?ll be able to cover all lines for every set of 7 sums. Look: we try to shift that set for ?The Well? ? 39 goes away, but we receive 53 in the other end, this set (41, 43, 45, 47, 49, 51) can fit ?Meeting? or ?Great Possessing? ? from totally ?young? (8-s and 7-s only) to totally ?old? (6-s and 9-s) hexagram. So we conclude that 39 and 53 must mark the same line. Analogically, we see other 4 pairs: 36-50, 37-51, 38-52 and 40-54. Without keeping it we won?t get access to all lines in all hexagrams.

Now we must arrange these numbers using those 2 rules. I simply put the most probable sum of 47 to the bottom ? for the initial line ? and distributed other sums to make the most even percents:

6th line: 44, 43, 40, 54
5th line: 49, 50, 36
4th line: 48, 51, 37
3rd line: 45, 52, 38
2nd line: 46, 39, 53
1st line: 47, 42, 41

Not very neat system at the first sight, yes, but it gives the following percents of probabilities:

6th: 13,85%
5th: 14,14%
4th: 16,86%
3rd: 15,06%
2nd: 18,78%
1st: 21,31%

Here the 1st line is also the most probable, but only 1,5 times more than 6th! Surely, one can restructure sums for lines to ensure top probabilities to say 2nd abd 5th lines and the lowest ? to 3rd and 6th lines.

If somebody is interested in full analysis with all figures, I can mail an MS Excel file with them. I didn?t want to overlaod you with all figures here. If you?d like, I can also calculate all these percents for 3 coins method.

Best regards (and sorry if I missed that original method in some earlier posts here),

Peter
 

peter

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Ouch, sorry, I used to look to the bottom of the page to see new posts, while they are already on the top
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peter

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Thanks, Wanderer, but personally I don't use text at all, I prefer non-textual methods, and they can deal with any amount of moving lines easily. But I thought that this stuff might be helpful here.
 

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