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In this topic I will empirically investigate the yarrow stalk method. My goal is to estimate the probabilities of the different types of lines. I have ordered yarrow stalks and they arrived today:

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I don't have yarrow stalks. I cut up barbecue skewers. I hope you enjoy interacting with the I Ching! Let us know what you figure out. • 1

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My first experiment will be estimating the variation of the number of stalks in the right and left heap.

In the above video found on YouTube the division into two heaps is done with eyes closed. Is that the usual way to do it?

Edited by wandelaar

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10 hours ago, wandelaar said:

In the above video found on YouTube the division into two heaps is done with eyes closed. Is that the usual way to do it?

Cannot find anything about it on the internet. Perhaps it doesn't matter as long as you just roughly divide the bundle in half and don't try to achieve a particular division.

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First results! I have repeatedly divided the bundle of 49 stalks just roughly in half as shown in the video but with my eyes open and counted the number of stalks in the right heap afterwards. I plan to do this a total of 200 times, but it is interesting to see the results so far:

28

27

22

23

25

22

24

26

21

27

27

22

23

23

24

28

27

25

29

24

27

24

27

25

23

30

24

24

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I ran the numbers above through a spreadsheet to calculate the first I Ching number.

 Count 28 Count of 2s 5 Count of 3s 23 Percent of 2s 17.86% Percent of 3s 82.14%

You'll recall that the only way to receive an initial 2 is with a sum of 9, e.g. 1+4+4. There are three ways to achieve a 3 (sum of 5), e.g. 1+3+1, 1+2+2, 1+1+3. So the textbook example says there is a 25% chance of finding an initial 2. However, from your numbers we see a 17.86% chance of finding an initial 2. This is an example of how the textbook percentages vary from the real world usage we had discussed in other threads.

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It could be that the variation in the number of stalks of the right heap differs when you divide bundles with less than 49 stalks. But that will come later. First I have to complete my research on the division of the bundle of 49 stalks.

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Results at last! I have performed the division in half of the bundle of 49 stalks for a total of a 100 times. I originally planned to do it 200 times, but I will keep it at a 100 times. I already had to repair one broken stalk, and I don't think more divisions will lead to a more precise estimations of the probabilities given that it's a human process where personal and circumstantial factors influence the precise manner of dividing the bundle of stalks. The division was done just roughly in half with open eyes, without trying to achieve any particular division. A few times (2 or 3) I redid the division when I saw even without counting that the division wasn't even roughly in half. I have counted the number of stalks after the divisions in the right heaps. See the uploaded files for the results.

Edited by wandelaar

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Here is first-pass analysis. It appears that 9% of the time you found a "2" value (9) and 91% of the time you found a "3" (5).

 RIGHT SIDE OFFSET ABS COUNT REMAINDER VALUE 28 4 4 0 19 3 3 27 3 3 1 27 2 3 22 -2 2 2 20 1 3 23 -1 1 3 16 2 3 25 1 1 4 9 0 2 22 -2 2 5 6 1 3 24 0 0 6 3 3 3

I've omitted the full list of numbers for brevity's sake.

Edited by Lost in Translation
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So there is quite some difference between the probabilities of the 2's and 3's found here, and the theoretical values from the books?

Is it possible to convert the date in the xls-file into a bar chart for the different numbers of stalks in the right heap?

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34 minutes ago, wandelaar said:

So there is quite some difference between the probabilities of the 2's and 3's found here, and the theoretical values from the books?

Is it possible to convert the date in the xls-file into a bar chart for the different numbers of stalks in the right heap?

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Here is a table of the frequencies for the number of stalks in the right heap:  frequencies.xls

Edited by wandelaar

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The diagram of the frequencies would have to approach a symmetrical bell curve when the division of the bundle of stalks is repeated many times over. Apparently even with a 100 divisions there still is a notable error-range. On the basis of the found results, the assumption of a bell curve, and the error-range I have estimated the ideal (= probabilistically most likely) outcome of the frequencies in column B of this file: estimation.xls

This is the corresponding diagram: • 1

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Now we can calculate the probabilities of the 9's and 5's on the basis of the ideal frequencies for a hundred divisions of the bundle of 49 stalks. Call the number of stalks in the right heap R and the number of stalks in the left heap L. Then we will only have to consider the cases where R = 19 to R=30 because other values of R don't happen. We will use the symbol mo(Z) for the result of repeatedly taking away 4 stalks from a heap of Z stalks until only 4, 3, 2 or 1 stalks are left. Then the number X (being 9 or 5) that results on the basis of the division of the bundle of 49 stalks can be found with the following formula:

X = 1 + mo(L) + mo(R-1) = 1 + mo(49-R) + mo(R-1)

Thus:

R=19   =>   X = 1 + mo(49-19) + mo(19-1) = 1 + mo(30) + mo(18) = 1 + 2 + 2 = 5   (with frequency   1)

R=20   =>   X = 1 + mo(49-20) + mo(20-1) = 1 + mo(29) + mo(19) = 1 + 1 + 3 = 5   (with frequency   2)

R=21   =>   X = 1 + mo(49-21) + mo(21-1) = 1 + mo(28) + mo(20) = 1 + 4 + 4 = 9   (with frequency   5)

R=22   =>   X = 1 + mo(49-22) + mo(22-1) = 1 + mo(27) + mo(21) = 1 + 3 + 1 = 5   (with frequency 11)

R=23   =>   X = 1 + mo(49-23) + mo(23-1) = 1 + mo(26) + mo(22) = 1 + 2 + 2 = 5   (with frequency 15)

R=24   =>   X = 1 + mo(49-24) + mo(24-1) = 1 + mo(25) + mo(23) = 1 + 1 + 3 = 5   (with frequency 16)

R=25   =>   X = 1 + mo(49-25) + mo(25-1) = 1 + mo(24) + mo(24) = 1 + 4 + 4 = 9   (with frequency 16)

R=26   =>   X = 1 + mo(49-26) + mo(26-1) = 1 + mo(23) + mo(25) = 1 + 3 + 1 = 5   (with frequency 15)

R=27   =>   X = 1 + mo(49-27) + mo(27-1) = 1 + mo(22) + mo(26) = 1 + 2 + 2 = 5   (with frequency 11)

R=28   =>   X = 1 + mo(49-28) + mo(28-1) = 1 + mo(21) + mo(27) = 1 + 1 + 3 = 5   (with frequency   5)

R=29   =>   X = 1 + mo(49-29) + mo(29-1) = 1 + mo(20) + mo(28) = 1 + 4 + 4 = 9   (with frequency   2)

R=30   =>   X = 1 + mo(49-30) + mo(30-1) = 1 + mo(19) + mo(29) = 1 + 3 + 1 = 5   (with frequency   1)

So 5 will happen 1+2+11+15+16+15+11+5+1 = 77 times, and 9 will happen 5+16+2 = 23 times. The probability of 5 is thus 77/100 = 0.77, and the probability of 9 is 23/100 = 0.23 .

Edited by wandelaar
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The second division of the bundle

For the second division of the bundle of stalks the 5 or 9 stalks found as a result of the first division are no longer used. So the second division is done with 49-5=44 or with 49-9=40 stalks. Because both are even numbers, the probability distributions of this second division will most likely have a "sharper peak" in the middle than we found for the probability distribution of the first division.

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I have now also performed the division in half of a bundle of 44 stalks for a total of a 100 times. The division was again done roughly in half with open eyes, without trying to achieve any particular division. I redid the division a few times when a stalk fell to the ground or when I saw even without counting that the division wasn't even roughly in half. I have counted the number of stalks after the divisions in the right heaps. See the uploaded files for the results.

Edited by wandelaar

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And here is the corresponding table of the frequencies for the number of stalks in the right heaps (for the division of a bundle of 44 stalks):   frequencies (44stalks).xls

Edited by wandelaar

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The diagram of the frequencies would have to approach a symmetrical bell curve with its top at 22 stalks when the division of the bundle of 44 stalks is repeated many times over. But just as in the case of a bundle of 49 stalks even with a 100 divisions there still is a big error-range. On the basis of the found results, the assumption of a bell curve with its top at 22 stalks, and the large error-range I have estimated the ideal (= probabilistically most likely) outcome of the frequencies in column B of this file:  estimation (44stalks).xls

And this is the corresponding diagram (for the division of the bundle of 44 stalks): Edited by wandelaar