About the probabilities of the traditional method
The yarrow stalk oracle is designed very cleverly to give the correct probabilities of 1,3,5,7 out of 16, but there is much dispute about the correctness of these values when the yarrow stalk method has been statistically analysed (eg. http://www.dentato.it/iching). But there is a logical flaw in their statistical analasys process which I'll try and explain with the analogy of using a books pages to pick random digits:
If there was a book which had 100 pages in it numbered from 00 to 99 and you decided to open it at random and use the page number to pick a random digit from 0-9, you would use the 1's digit, not the 10's digit because it changes on every page and would therefore be more random. All would agree that each digit would have a 1/10 chance of occuring when randomly opening the book.
But what if we do the same experiment with a 95 page book instead? What would the probability of each digit be now? It's still 1/10 for each because our random pick is not a uniform probability distribution over all pages, but rather a normal disribution with it's mean in the middle of the book - nobody would naturally pick the pages right at the ends. But because of the 1's digit changing on every page the result is very good randomness when we open the book in a natural way.
In the traditional yarrow stalk method, using remainders after counting by fours plays exactly the same role as the unpredictable 1's digit in the preceeding book analogy. It's a process that's designed to yield a uniformly distributed result from a normally distributed selection. It's designed for the splitting of the pile to be a natural and vaguely central division. The total number of stalks in the pile does not affect the probability as long as the initial amount is fairly high - and 49 is high enough to give quite stable results.
The Divination Process
The yarrow stalk method relates the divination process to the energy of the time and also to the King Wen arrangement. Here is an extract from the Wilhelm translation of the I Ching regarding the yarrow stalk divination method (From chapter IX of Ta Chuan).
| One takes fifty yarrow stalks, of which only forty-nine are used. These forty-nine are first divided into two heaps (at random), then a stalk from the right-hand heap is inserted between the ring finger and the little finger of the left hand. The left heap is counted through by fours, and the remainder (four or less) is inserted between the ring finger and the middle finger. The same thing is done with the right heap, and the remainder inserted between the forefinger and the middle finger. This constitutes one change. Now one is holding in one's hand either five or nine stalks in all. The two remaining heaps are put together, and the same process is repeated twice. These second and third times, one obtains either four or eight stalks. The five stalks of the first counting and the four of each of the succeeding countings are regarded as a unit having the numerical value three; the nine stalks of the first counting and the eight of the succeeding countings have the numerical value two. When three successive changes produce the sum 3+3+3=9, this makes the old yang, i.e., a firm line that moves. The sum 2+2+2=6 makes old yin, a yielding line that moves. Seven is the young yang, and eight the young yin; they are not taken into account as individual lines.
|
If we make a table of different splits of the forty-nine over an arbitrary range, say twenty-one and twenty-eight to thirty and nineteen, then we can see the results as a repeating sequence. The sequence shows that nines are only one-in-four, and the rest fives. However the two subsequent changes start with an even amount, and this leads to half fours and half eights. The table below on the left shows a first change from forty-nine, the right table shows an example of a second or third change starting at forty. The columns for each heap show the total of the heap and the remainder after counting through by fours; the right heap always has one of these removed, so the result is the remainders plus one.
The fours and fives form units of numerical value three, the eights and nines of value two. Changes are therefore composed of three digits, each a two or a three. The first digit (the one derived from the first change) has only 25% chance of being a two but 75% toward three, the other two digits are 50/50.
| Left Heap
|
Right Heap
|
Result
|
| 21 |
20,1 |
28 |
24,3 |
5
|
| 22 |
20,2 |
27 |
24,2 |
5
|
| 23 |
20,3 |
26 |
24,1 |
5
|
| 24 |
20,4 |
25 |
20,4 |
9
|
| 25 |
24,1 |
24 |
20,3 |
5
|
| 26 |
24,2 |
23 |
20,2 |
5
|
| 27 |
24,3 |
22 |
20,1 |
5
|
| 28 |
24,4 |
21 |
16,4 |
9
|
| 29 |
28,1 |
20 |
16,3 |
5
|
| 30 |
28,2 |
19 |
16,2 |
5
|
Heaps divided from 49 stalks
|
| Left Heap
|
Right Heap
|
Result
|
| 21 |
20,1 |
19 |
16,2 |
4
|
| 22 |
20,2 |
18 |
16,1 |
4
|
| 23 |
20,3 |
17 |
12,4 |
8
|
| 24 |
20,4 |
16 |
12,3 |
8
|
| 25 |
24,1 |
15 |
12,2 |
4
|
| 26 |
24,2 |
14 |
12,1 |
4
|
| 27 |
24,3 |
13 |
8,4 |
8
|
| 28 |
24,4 |
12 |
8,3 |
8
|
| 29 |
28,1 |
11 |
8,2 |
4
|
| 30 |
28,2 |
10 |
8,1 |
4
|
Heaps divided from 40 stalks
|
The following list of sixteen changes represents the correct probabilities of each digit. Out of the sixteen there are only four starting with a two, but half having two in the second or third places. Counting up the old and young yin and yang changes reveals the probabilities in the table below.
| 2,2,2 |
x
|
3,2,2 |
|
3,2,2 |
|
3,2,2 |
|
| 2,2,3 |
|
3,2,3 |
|
3,2,3 |
|
3,2,3 |
|
| 2,3,2 |
|
3,3,2 |
|
3,3,2 |
|
3,3,2 |
|
| 2,3,3 |
|
3,3,3 |
o
|
3,3,3 |
o
|
3,3,3 |
o
|
|
Yarrow Approximation with Four Coins
As can be seen from the second column of the table below the probabilities don't match the traditional three coin method. By using four coins however, they can represent a binary number from zero to fifteen. By arranging the coins in a line and assigning them the values one, two, four and eight, and then adding the values of all the coins which have heads up. The third column of the table below shows how to convert this resulting sum to a change.
| Change |
Probability |
Sum
|
x |
1/16 |
0
|
o |
3/16 |
1-3
|
|
5/16 |
4-8
|
|
7/16 |
9-15
|
|
It doesn't matter how you like the coins to be arranged in shape, or which ones get assigned with which values. Find the way you're most comfortable with, for example all in a line assigning one, two, four and eight left to right. Or maybe in a square with one and two on top, and four and eight at the bottom. The key is consistency, once you've found a way which is natural and easy to remember, stick with it. Readings will become clearer as attention can remain more on the situation and the time, and less on the process of divination itself. Here's a table of all the possible coin thows and their corresponding change.
| Coins |
Change
|
| T T T T |
x
|
| T T T H |
o
|
| T T H T |
o
|
| T T H H |
o
|
|
| Coins |
Change
|
| T H T T |
|
| T H T H |
|
| T H H T |
|
| T H H H |
|
|
| Coins |
Change
|
| H T T T |
|
| H T T H |
|
| H T H T |
|
| H T H H |
|
|
| Coins |
Change
|
| H H T T |
|
| H H T H |
|
| H H H T |
|
| H H H H |
|
|
|
See also
