The traditional method of consulting the I Ching is by using the stalk oracle. This is done by performing a procedure which uses 50 yarrow stalks.
Yarrow is a common plant which is native to many countries. Its botanical name is Achilles Millefolium. The name is reported to originate partly from the fact that Achilles used it to staunch the bleeding wounds of his soldiers. The Millefolium part of the name comes from the many segments of its foliage. Other names for the plant are:
Milfoil
Soldier's Woundwort
Thousand Weed
Nose Bleed
Bloodwort
Staunchweed
Sanguinary
and many others. Its foliage, as well as its stalks, have been used for the purpose of divination.
There are several rituals associated with the stalk oracle, including the burning of incense, the proper storage and handling of the stalks, and the proper mental preparation. My purpose here is to address the mathematics of the process, not the traditions. The traditions and their rationale are handled in great detail in the references.
First, I will describe the procedure. Then I will analyze the probabilities associated with it.
The oracle uses 50 yarrow stalks. At the beginning, one of the stalks is removed and placed aside. It is not used during the procedure. This leaves 49 stalks.
The 49 stalks are used to obtain the six lines of the hexagram. This is done by counting out the stalks to obtain a numerical value of either 2 or 3, and repeating the counting procedure three times for each line. I describe the details about how the stalks are counted later on.
Once the stalks have been counted, the seeker has three numerical values, each of which has the value of 2 or 3. The three values are added to obtain the numerical value of the line. You should be able to see the similarity with the coin oracle. Each line of the hexagram is obtained by adding the following values:
(2 or 3) + (2 or 3) + (2 or 3)
There are eight possible summations using these numbers:
2 + 2 + 2 = 6
2 + 2 + 3 = 7
2 + 3 + 2 = 7
2 + 3 + 3 = 8
3 + 2 + 2 = 7
3 + 2 + 3 = 8
3 + 3 + 2 = 8
3 + 3 + 3 = 9
The summation results in one of four possible values, which are 6, 7, 8, or 9. The even numbers represent yielding lines, and the odd numbers represent firm lines. The numerical values are mapped to lines using the following:
6 moving yielding line
7 static firm line
8 static yielding line
9 moving firm line
So far, the numerology is very similar to the coin oracle.
Each line is derived by using the stalks three times and summing the results of the three draws. Each draw of the number 2 or 3 requires counting out the stalks three times and summing the results to obtain either the 2 or the 3. Therefore, the entire hexagram requires the derivation of 18 numbers of value 2 or 3, each derivation of a 2 or a 3 requires three uses of the stalks. This amounts to the use of 54 countings of the stalks to derive the entire hexagram. This will become more clear as I describe the details of the procedure.
This section describes the details of how the stalks are counted to derive a single line of the hexagram. This is a somewhat involved procedure, so please bear with me.
The 49 stalks are split into two roughly equal piles. Then, one stalk is taken from the right pile and placed between the little finger and the ring finger of the left hand. This leaves 48 stalks remaining in the two piles. The left pile is picked up in the left hand and the stalks are removed four at a time and placed aside until only one to four stalks remain in the left hand. These 1 to 4 stalks are placed between the ring finger and the middle finger of the left hand. The stalks which were counted out of the left pile are temporarily set aside.
The stalks in the right hand pile are now picked up in the left hand. Once again, the stalks are counted out four at a time until 1 to 4 stalks remain in the left hand. The stalks which were removed are placed aside. The 1 to 4 stalks are placed between the middle finger and the index finger of the left hand. Now, all the stalks in the left hand are counted. As it turns out (we will discuss why later) only two numbers of stalks will remain in the left hand after this procedure, either 9 or 5.
At this point in the procedure, the one stalk held between the little finger and the ring finger is "disregarded as supernumerary" (from the R. Wilhelm translation). The single stalk is set aside and the sum is changed to either 8 or 4. These numbers are then mapped into two numbers to be used in the construction of the hexagram. The sum of 4 is mapped into the number 3. The sum of 8 is mapped into the number 2. The metaphysical reasons for these mappings are beyond the scope of this discussion, and I must once again refer you to the references. Let it suffice to say that the numbers 2 and 3 contain the numerically important significance, and the summations 4 and 8 are merely the procedural mechanism for deciding which of the two significant numbers has been obtained for the oracle.
The number which was just obtained (2 or 3) is noted, and the stalks in the left hand are placed aside.
All of the stalks which were counted out of the left hand (from both piles) are again combined into a single pile. The stalks which remained in the left hand (and numbered either 9 or 5) are not included in the next step. They are set aside.
The remaining stalks are again divided roughly into two equal piles. One stalk is again taken from the pile on the right side and placed between the little finger and the ring finger of the left hand. Then the left pile is picked up in the left hand. The stalks are counted out by fours until only 1 to 4 stalks remain in the left hand. These are placed between the ring finger and the middle finger of the left hand. The same is done with the stalks from the right pile, placing the 1 to 4 remaining stalks between the middle finger and the index finger of the left hand. When the stalks in the left hand are summed, they will add up to either 8 or 4. Once again, the reason for this will be discussed in the following analysis.
Note that the single stalk held between the little finger and the ring finger of the left hand is not disregarded this time.
Once again, the 8 is mapped into a 2 and the 4 is mapped into a 3. The derived number (2 or 3) is noted and the procedure is continued. The stalks remaining in the left hand are set aside and not used in the next step.
The third use of the stalks is identical with the second use of the stalks. It also provides either a 2 or a 3.
Once the three derivations have been made, the line is determined by adding up the derived values to obtain the value which represents the line. The following chart contains all the possible combinations:
2 + 2 + 2 = 6
2 + 2 + 3 = 7
2 + 3 + 2 = 7
2 + 3 + 3 = 8
3 + 2 + 2 = 7
3 + 2 + 3 = 8
3 + 3 + 2 = 8
3 + 3 + 3 = 9
The resulting numbers come from the set {6, 7, 8, 9} with varying probabilities. The line itself is obtained by mapping the above set of numbers into the lines using the following table:
6 is a moving yielding line
7 is a static firm line
8 is a static yielding line
9 is a moving firm line
The entire procedure described above is performed six times to obtain the six lines of the hexagram. The determination of each line begins with 49 stalks. The lines are determined from the bottom up. Thus, the first line is the bottom line and the sixth line is the top line.
While this procedure is difficult to explain in words, it is actually quite simple in practice. Much of it is repetitive.
I will now analyze the probabilities of the resulting numbers.
The procedure begins with 50 stalks. One of those is removed and not used. Hence, the real count for oracle is 49 stalks. I have not found any reason for having 50 stalks yet, so I must assume that it is a long tradition with a probable source in numerology.
The first act is to divide the stalks into two roughly equal piles. This is where the randomisation occurs. A somewhat random number of stalks will be in each pile depending upon how the stalks are divided. This step is equivalent to the toss of a coin in the coin oracle. At this point in the procedure, the resulting line has actually been determined.
One stalk is removed from the right hand pile and placed between the little finger and the ring finger of the left hand. However, careful study of the procedure shows that eventually, when the stalks in the left hand number either 9 or 5, the single stalk is disregarded to make the count either 8 or 4. This is unique to the first counting of the stalks. In the last two countings, none of the stalks is disregarded, and the count results in either 8 or 4 as well. This fact will prove significant in subsequent discussions.
Since we have removed two stalks from the original 50, only 48 stalks remain in the two piles. So, the first count begins with 48 stalks divided into roughly two equal piles and the 49th stalk held between the little finger and the ring finger of the left hand. To analyze the procedure we will define the following names:
T is the total number of stalks (48)
L is the number of stalks in the left pile
R is the number of stalks in the right pile.
The left pile has 1 to 47 stalks in it. We know that the right pile contains the remaining stalks, so we can say:
R = T - L
which basically means that the number of stalks in the right pile is 48 minus the number of stalks in the left pile. This number also has a range of 1 to 47, however it is related to the number of stalks actually in the left pile by the equation above.
The significant fact about the number 48 is that it is evenly divisible by 4. This is significant because we will be counting out the stalks by 4. As we count out the stalks, we will eventually end up with 1, 2, 3, or 4 stalks in the left hand. If there are any more stalks than that, say 5, we will count out another group of 4 before we quit (down to 1 in the example). Therefore, the only possible outcomes for this part of the procedure are the four numbers, {1, 2, 3, 4}. The reason that we have stalks remaining in the left hand is that, even though the 48 stalks will always be evenly divisible by 4, the number of stalks in the left hand pile is not necessarily divisible by four.
Now we proceed to count out the right hand pile of stalks. This pile also has an unknown number of stalks in it, so when we count out the stalks in the right hand pile we will, once again, obtain one of the numbers {1,2,3,4}. But there is a difference this time. The number of stalks in the right hand pile is not independent of the number of stalks in the left hand pile. In fact, they are related by the equation above. If you take some example counts, you will see that, because the total number of stalks is evenly divisible by 4, the results of the counts are restricted to the following outcomes:
Left Outcome Right Outcome Summation 1 3 4 2 2 4 3 1 4 4 4 8
Why is this? Well, the total number of stalks (in this case, 48) is evenly divisible by 4. Therefore, counting out all of the stalks (which we have done by counting out both piles) must result in a final count of 4, or zero if you count out that final 4 stalks. But we did not count out the final 4 stalks, some of them remain from the first count, and some of them remain from the second count. By splitting the 48 stalks into two separate piles to count, we have left some of the final 4 stalks in the left hand from first count and the rest of them from the second count. The mathematics therefore requires that the sum of the two separate counts must add up to 4. If you look at the table above, you will notice that the two outcomes always add up to 4.
Well, almost.
Where did the 8 come from? When we counted out the stalks from each pile, we did not count out the last 4 stalks, but kept them in the left hand. Hence, we have two groups of 4 stalks left over instead of one. If we counted out those last two groups of 4 stalks we would end up with zero stalks because the number 48 is evenly divisible by 4. Once again, this outcome is the result of counting out the 48 stalks from two piles instead of one.
This long discussion has revealed why we will always have either 4 or 8 stalks remaining when beginning with 48. The number 48 will always count down to a whole number of groups of 4 stalks. There will always be one group of 4 stalks left from both counts, or two groups of 4 stalks from both counts. Which result you get is determined by the number of stalks in the left pile in relation to the number of stalks in the right pile. If the left pile and the right pile have numbers which are evenly divisible by 4, you will end up with two groups of 4 stalks left over. A little further thought reveals that this will be true of any beginning number of stalks which is evenly divisible by 4.
This first count is unique because we still have that 49th stalk in the left hand. Let's look at the second count.
The oracle requires that we put aside the stalks remaining in the left hand prior to proceeding to the next step. Since we have 49 stalks total, and we have either 5 or 9 stalks in the left hand, the second step of the oracle will begin with either 44 or 40 stalks. Each of these numbers is evenly divisible by 4, so the same rules apply to the second count as applied to the first count. But note that during the second count we do not remove one stalk from the right hand side and discard it as we did in the first count. The single stalk is placed into the left hand and counted later. Thus the total number of stalks is still divisible by 4, and since we are eventually counting out all of the stalks, the resulting stalks in the left hand must add up to 4 or 8 even though they are now in three groups (the single stalk, the left hand count, and the right hand count).
This shifts the numbers in the table to be:
Single Stalk Left Outcome Right Outcome Summation 1 1 2 4 1 2 1 4 1 3 4 8 1 4 3 8
Notice that for the second count the table shows two sums of 4 and two sums of 8. The first count resulted in three sums of 4 and only one sum of 8. So the probabilities for the second count are different than they are for the first count.
The third count is subject to the same rules as the second count, but it begins with a different number of stalks. The second count begins with either 44 or 40 stalks and we will set aside either 4 or 8. So the third count will begin with:
44 - 4 = 40
or
44 - 8 = 36
or
40 - 4 = 36
or
40 - 8 = 32
These are all of the possible numbers with which the third count will begin. All of these numbers are evenly divisible by 4, hence all of them will result in a sum of either 4 or 8 stalks remaining in the left hand.
To summarize, the procedure of the oracle reduces to three counts of the stalks, each resulting in a value of 4 or 8. The values of 4 or 8 are then mapped directly into the values 3 or 2 and summed. This is similar to the coin oracle, which also sums three numbers of value 3 or 2.
But are the probabilities the same as the coin oracle?
The table of possible outcomes for the first count is:
Left Outcome Right Outcome Summation 1 3 4 2 2 4 3 1 4 4 4 8
You will notice that three out of the four outcomes result in the number 4, and only one results in the number 8. Therefore, we can assign the following probabilities to the two numbers:
4 has probability of 0.75
8 has probability of 0.25
It is three times more probable to draw a 4 as it is to draw an 8. Since the values of 4 and 8 are mapped directly into the values 3 and 2, we can say:
3 has probability of 0.75
2 has probability of 0.25
For the second and third counts, the table is:
Single Stalk Left Outcome Right Outcome Summation 1 1 2 4 1 2 1 4 1 3 4 8 1 4 3 8
Mapping 4 to 3 and 8 to 2, we can assign the probabilities:
3 has probability of 0.5
2 has probability of 0.5
From this, we can calculate the probabilities of all the eight possible outcomes for the oracle. The eight outcomes are:
2 + 2 + 2 = 6
2 + 2 + 3 = 7
2 + 3 + 2 = 7
2 + 3 + 3 = 8
3 + 2 + 2 = 7
3 + 2 + 3 = 8
3 + 3 + 2 = 8
3 + 3 + 3 = 9
You must keep in mind during this calculation that the probabilities for the first column of numbers is different from the probabilities for the second and third columns of numbers. We can multiply the probabilities together to obtain:
2 + 2 + 2 = 6 0.25 x 0.5 x 0.5 = 0.0625
2 + 2 + 3 = 7 0.25 x 0.5 x 0.5 = 0.0625
2 + 3 + 2 = 7 0.25 x 0.5 x 0.5 = 0.0625
2 + 3 + 3 = 8 0.25 x 0.5 x 0.5 = 0.0625
3 + 2 + 2 = 7 0.75 x 0.5 x 0.5 = 0.1875
3 + 2 + 3 = 8 0.75 x 0.5 x 0.5 = 0.1875
3 + 3 + 2 = 8 0.75 x 0.5 x 0.5 = 0.1875
3 + 3 + 3 = 9 0.75 x 0.5 x 0.5 = 0.1875
Once again, we have a situation where more than one calculation results in a specific line of the hexagram. To calculate the total probability for any specific line to be drawn, we add up the probabilities of all the sums which will result in that line. Therefore, we obtain the following table:
6 has probability = 0.0625
7 has probability = 0.0625 + 0.0625 + 0.1875 = 0.3125
8 has probability = 0.0625 + 0.1875 + 0.1875 = 0.4375
9 has probability = 0.1875
It is interesting to compare these numbers to the numbers from the coin oracle. The coin oracle numbers are:
6 (moving yin line) 0.125
7 (static yang line) 0.375
8 (static yin line) 0.375
9 (moving yang line) 0.125
You may note that for the coin oracle, the probability of a static yin line or a static yang line are equal (0.375). Also the probability of a moving yin line or a moving yang line are equal (0.125). This is not the case for the stalk oracle. The probabilities for the stalk oracle are, in decreasing order:
8 (static yin line) 0.4375
7 (static yang line) 0.3125
9 (moving yang line) 0.1875
6 (moving yin line) 0.0625
I have sorted the above list to show which lines are more probable.
So, why are the probabilities different for the stalk oracle? The fly in the ointment, if you will, is the use of the 49th stalk on the first count. This single stalk is disregarded after the count, however it forces the seeker to count out the remaining 48 stalks into two groups (the one from the left pile and the one from the right pile). In the second and third count, the single stalk is not disregarded, so the stalks end up in three groups (the single stalk plus those from the left pile and those from the right pile). The removal of the single stalk into its own group in the second and third case causes the probabilities of a 2 and a 3 to be different from the probabilities of the first count.
In the coin oracle, each coin is the equivalent of one of the counts of the stalks because each coin determines a 2 or a 3 when it is tossed. However, each coin has an equal probability of returning a 2 or a 3. That is not the case with the stalk oracle. For the second and third count of the stalks, each count has equal probability of returning a 2 or a 3, but the first count provides a higher probability of returning a 3 than a 2. Hence, the ultimate probabilities for the stalk oracle are different than they are for the coin oracle.
The stalk oracle could be modified to provide the same probability distribution as the coin oracle by removing the 49th stalk along with the 50th stalk and simply not using it at all. Then, if the procedure for the second count and the third count were used for the first count as well, the probability of a 3 and a 2 would be identical in the first count. This is the case for the coin oracle, all the coins being equal.
The coin oracle could be modified to emulate the stalk oracle. Take two of the coins and toss them. If both yin sides (the side with the inscription) are up, choose a 2. Otherwise choose a 3. This would provide the first value for the sum. Notice that with two coins tossed, the following outcomes are possible:
yin yin
yin yang
yang yin
yang yang
There are three cases where at least one yang side is up and only one case in which both sides are yin. Therefore, the probabilities of yin (2) and yang (3) values are:
yang (3) has probability 3 in 4 = 0.75
yin (2) has probability 1 in 4 = 0.25
These are the same as for the first count of the stalk oracle. Therefore, we could use this first toss of two coins to determine the first value for the line. Then, the two coins could be tossed again, and each side of each coin would determine one of the other two values for the line. When tossing a single coin to determine a 2 or a 3, there is an equal chance for each of the values. Therefore, tossing two fair coins for the second and third values carries the same probabilities as the second and third count of the stalk oracle.
These modifications to the oracles are pure speculation by me, and are intended only to illuminate the underlying mathematics. I am not suggesting that we contradict ancient tradition by modifying the oracles.
So, using the newly calculated probabilities for the stalk oracle, let's look at some interesting hexagrams and their probabilities.
In our discussion of the coin oracle we calculated the probability of drawing any single hexagram with all static lines and compared that with the probability of drawing the same hexagram with some moving lines. We can now duplicate that calculation for the stalk oracle. As you recall, we begin by calculating the probability of drawing any specific hexagram regardless of whether or not any of the lines move.
A problem appears to arise now because the probabilities of the individual types of lines are no long equal as shown in the following chart:
8 (static yin line) 0.4375
7 (static yang line) 0.3125
9 (moving yang line) 0.1875
6 (moving yin line) 0.0625
However, you will notice that if you add up the probabilities of drawing any type of yin line (moving or static) they add up to 0.5. The same is true for the yang lines:
0.4375 + 0.0625 = 0.5
0.3125 + 0.1875 = 0.5
Therefore, the probability of drawing a yin or a yang of any type (moving or static) is still equal, as it was in the coin oracle. Hence, we can conclude that the drawing of any specific hexagram regardless of the movement of the line is still 1/64, or 0.015625.
We can then calculate the probability of drawing that hexagram with all the lines static. However, the result here must be different than it is for the coin oracle because the relative probability of drawing a static yin line is different from the probability of drawing a static yang line. So, for example, to calculate the probability of drawing the Creative:
with no moving lines we must multiply together six times the probability of drawing a single static yang line:
0.3125 x 0.3125 x 0.3125 x 0.3125 x 0.3125 x 0.3125 = 0.0009313225746155
This differs from the probability of drawing a static Receptive hexagram:
which must use the probability of drawing a single static yin line:
0.4375 x 0.4375 x 0.4375 x 0.4375 x 0.4375 x 0.4375 = 0.00701242685318
Hence, you can see that the probability of drawing the Creative hexagram with at least one moving line is just the probability of drawing the hexagram regardless of the motion of the lines minus the probability of drawing that same hexagram with all the lines static.
0.015625 - 0.0009313225746155 = 0.01469367742539
Similarly, the probability of drawing the Receptive with at least one moving line is:
0.015625 - 0.00701242685318 = 0.00861257314682
As you can see, these are slightly different results. In the case of the coin oracle, these two probabilities were identical.
Further calculations are left up to the reader. Perhaps some more interesting comparisons can be made between the two oracles.
