# Cross the Great Stream

Investigations into the I Ching

### Part I

It’s obvious to anyone who’s tried it that the yarrow-stalk method of consulting I Ching is superior in terms of the production of a meditative state through the routine automatism of the fairly long and convoluted process of drawing stalks. The result is ‘seeded’ right at the beginning, in the sense that you could count the number of stalks in the two piles and immediately tell the result. However the long drawing-out of the consequences of the random/unconscious division of the stalks seems as important; there’s an interesting sense in which the automatism of following the trail seeded by the division, in turn ‘seeds’ the contemplative state necessary to make the next division.

Anyway, I’d wondered for a long time about the probabilities involved: it seems obvious that the stalks are a more complex procedure than coins and that the chances of getting the different lines are not identical. Luckily, before being tempted into what would inevitably have been another lengthy bout of painful number-crunching, I found a summary which shows the probabilities as follows:

``` -x- Yarrow : 1/16 Coins : 2/16
--- Yarrow : 5/16 Coins : 6/16
- - Yarrow : 7/16 Coins : 6/16
-o- Yarrow : 3/16 Coins : 2/16```

Obviously, the stalks are far more ‘weighted’, and arguably more interesting for it; the slight disparity between the 7 and 8 (‐‐‐ and ‐ ‐) lines, and the radical disparity in the chances of getting the two moving lines.

It’s easy to work out that (not including moving lines, that’s too much to do by hand!) there are four composite probabilities of getting the trigrams : 125/4096 (all 7s), 343/4096 (all 8s) 245/4096 (two 7s one 8) and 175/4096 (two 8s one 7)

```___
___
___

Chhien = 5/16 * 5/16 * 5/16 = 125/4096 001111101 / 100000000000
```
```- -
- -
- -

Khun = 7/16 * 7/16 * 7/16 = 343/4096 101010111 / 100000000000
```
```- -
- -
---

Chen = 7/16 * 7/16 * 5/16 = 245/4096 011100001 / 100000000000```
```- -
---
- -

Khan = 7/16 * 5/16 * 7/16 = 245/4096 011100001 / 100000000000```
```---
- -
- -

Kên = 5/16 * 7/16 * 7/16 = 245/4096 011100001 / 100000000000```
```---
---
- -

Sun = 5/16 * 5/16 * 7/16 = 175/4096 010101111 / 100000000000```
```---
- -
---

Li = 5/16 * 7/16 * 5/16 = 175/4096 010101111 / 100000000000```
```- -
---
---

Tui = 7/16 * 5/16 * 5/16 = 175/4096 010101111 / 100000000000
```

The composite probabilities of the hexagrams, then:

```Chhien = 125/4096 * 125/4096 = 15625 / 16777216
Khun = 343/4096 * 343/4096 = 117649 / 16777216
Chun = 175/4096 * 175/4096 = 30625 / 16777216
Mêng = 175/4096 * 175/4096 = 30625 / 16777216
Hsü = 175/4096 * 245/4096 = 42875 / 16777216
Sung = 125/4096 * 175/4096 = 21875 / 16777216
Shih = 343/4096 * 175/4096 = 60025 / 16777216
Pi = 343/4096 * 175/4096 = 60025 / 16777216
Hsiao Hsü= 245/4096 * 125/4096 = 30625 / 16777216
Li = 125/4096 * 245/4096 = 30625 / 16777216
Thai = 343/4096 * 125/4096 = 42875 / 16777216
Phi = 125/4096 * 343/4096 = 42875 / 16777216
Thung Jen= 125/4096 * 245/4096 = 30625 / 16777216
Ta Yu = 245/4096 * 125/4096 = 30625 / 16777216
Chhien = 343/4096 * 175/4096 = 60025 / 16777216
Yü = 175/4096 * 343/4096 = 60025 / 16777216

[....TBC.....]
```

By pairing up the trigrams, you’d expect there to be more distinct probabilities, but surprisingly there are fewer because 175*175=245*125. But you can see that pairing up the trigrams results in huge disparities, of an order of magnitude, in the likelihood of getting (to take the worst instance) Chhien and Khun. More generally the slightly smaller probability of getting a broken line means that the more a hexagram is dominated by these receptive/female lines the less likely it is to turn up (this takes no account of the different meanings of the relative positions of the lines, of course).

### Part II

Thinking about the process involved in the yarrow-stalk method of I Ching divination led to a problematising of the provenance of the probabilities given for each line:

```-x- 1/16 (0.0625)
--- 5/16 (0.3125)
- - 7/16 (0.4375)
-o- 3/16 (0.1875)
```

Given that the process begins with the person consulting the I Ching dividing the stalks, thus presumably providing the basis for the end result, it makes sense to ask whether any systematic factor (that is, any factor other than the specific number of stalks in each of the divided piles, which numbers are analytically tied to a specific result) could be identified as having an effect on the probabilities of obtaining the different types of line. It didn’t seem at all obvious that however unequally the pile was divided the probability distribution would remain identical.

The experiment described below suggests that the division of the stalks ‘into two equal piles’ represents a ‘perfect model’ or limit of the system, a limit, moreover, from which the system has to stray in order for it to ‘work’ (the constitutive importance of human error, noise or ‘randomness’). Furthermore, it shows how the tolerated range of this straying from the ‘perfect model’ affects the probability distributions, and ultimately, the judgments obtained through this method of divination. Although ‘noise’ plays a random-like role, the amount of noise as a measurable quantity is correlated with definite changes in the system.

### Algorithmic Model

The random seeding takes place with the removal and discarding from the entire pile of 50 stalks P of one stalk and the splitting of the remainder into two piles: so the initial domain is

```all (a,b) where (a+b=P)
```

or, I suggest,

```all (a,b) where (a+b=P and (.5P-x)<a<(.5P+x)))
```

Where x is the limiting factor of how differently-sized the two piles may be (how far either may diverge from P/2 or half of the original pile size) before they are no longer to be realistically considered an intuitive ‘half-and-half’ split. I would suggest that rather than a constant, x should be proportional to the pile size, so:

```all (a,b) where (a+b=49 and (.5P-(P/x))<a<(.5P+(P/x)))
```

Whether or not the variation of x has any bearing on the probability distribution of the final result is an interesting question in regard to the possibility of the person consulting the oracle’s subconsciously influencing the result (say, if the more equal the piles were, the more likely one was to obtain moving lines, or something of that nature).

One of the stalks from the right pile is ‘stored’ in the left hand, following which the left pile is counted out modulo 4 (four at a time, with the remainder, <=4, being stored). At this time the total of ‘stored’ stalks is then:

```1 + a mod 4
```

[I’m aware that this is a misuse of ‘mod’ – if anyone can suggest the right notation let me know]

Next the right pile is similarly counted out and the remainder stored:

```1 + a mod 4 + (b-1) mod 4
```

Where the possibilities are:

```1+4+4, 1+2+2, 1+1+3
```

so either 9 or 5, with a 2/3 chance of 5.

In this first count, with its extra stalk, 9 counts for 8, 5 for 4: so

```r1 = (1 + a mod 4 + (b-1) mod 4 ) -1
```

The tricky part is that 4 is considered a ‘whole’, thus it is ultimately counted as 3, and 8 a ‘double’, counted as 2. But we will deal with this later.

Now we divide the remaining stalks again, giving a member of :

```all (a,b) where (a+b=(49-r1) and (.5P-(P/x))<a<(.5P+(P/x)))
```

and once again:

`r2 = (1 + a mod 4 + (b-1) mod 4 ) -1`

and a third time, where we yield a member of:

```all (a,b) where (a+b=(49-r1-r2)) and (.5P-(P/x))<a<(.5P+(P/x)))
```

ending up with:

`r3 = (1 + a mod 4 + (b-1) mod 5 ) -1 `

Our three results will be:

`r1-1 = 8 or 4 r2 = 8 or 4 r3 = 8 or 4`

which together make up one line (which can be 6,7,8, or 9):

```12 (4+4+4) = (3+3+3) => 9
16 (8+4+4) = (2+3+3) => 8
20 (8+8+4) = (2+2+3) => 7
24 (8+8+8) = (2+2+2) => 6```

The order reversal (the smaller values becoming the larger) arises from the intermediate translation of 4 to 3 (“unity”) and 8 to 2 (“double”) which is an axiom of the system. The same result can easily be obtained as follows

`12 - (((r1-1)+r2+r3) / 4)`

So, given that each subsequent line will be probabilistically independent, and assuming that x remains constant through the three draws (not unreasonable since presumably we would attribute x to the person consulting the oracle) our question will be: what is the probability of obtaining 6,7,8 or 9 with the following algorithm:

```F = 12 - ((r1-1)+r2+r2 / 4)
given (P,x,a,b,a',b',a'',b'',r1,r2,r3)
where:
a+b=P and (.5P-P/x)<a<(.5P+P/x),
r1 = (1 + a mod 4 + (b-1) mod 4),
a'+b'=(49-r1) and
(.5(P-r1))-(P-r1/x))<(P-r1)<(.5(P-r1)+(P-r1/x)),
r2 = (1 + a' mod 4 + (b'-1) mod 4 )
a''+b''=(49-r1-r2)) and
(.5(P-r1-r2)-(P-r1-r2/x))<(P-r1)<(.5(P-r1-r2)+(P-r1-r2/x)),
r3 = (1 + a'' mod 4 + (b''-1) mod 4 )```

### Testing the New Model

Testing this for every combinatorial possibility of a,b,a’,b’,a”,b”, for different values of x would therefore yield the probabilities for each line for each x, from which we could proceed to the combined probabilities for whole hexagrams.

A perl program was written to sum results from all combinatorial possibilities given x. As might be expected, the results show that the initial division of the pile acts as the random ‘seed’; this done, the outcome is fixed. However the real question was whether the probabilities would be as stated.

An initial run was tried with x=16, meaning that the size of the divided piles could not vary more than P/16 from P/2 – in the initial division this means that neither pile can be less that (49/2)-(49/16)=24.5-3=21.5 or more than (49/2)-(49/16)=24.5-3=27.5, giving a total of 6 possible divisions (22/27, 23/26, 24/25, 25/24, 26/23 and 27/22). With the disparity of the piles thus limited, the probabilities of obtaining the different lines, expressed in percentages and probabilities, are as follows:

```6 Old Yin      - x - 3.157 (0.3157) - suggested (0.0625)
7 Young Yang   ----- 38.421 (3.8421) - suggested (0.3125)
8 Young Yin    -- -- 39.473 (3.9473) - suggested (0.4375)
9 Old Yang     - o - 18.947 (1.8947) - suggested (0.1875)
```

As can be seen, the relationship between these probabilities follows proportionally that suggested, but the fit is by no means exact. This once again returns us to the intuition that the role of x is non-trivial, and that the probabilities given were the result of fixing x.

Modifying the program to iterate x from 20 down to 2 (that is, from P/20 to P/2) yielded a matrix of the probabilities of obtaining each type of line given these different values of x. The results are surprising. Although some noise could be expected in the progression, due to the effect of rounding to the “nearest stalk”, the graph below shows clear divergences as x drops.

[graph lost in the mists of time, sorry]

Of course, our limit case of x=2 is unrealistic (we can safely assume that no-one, asked to split a pile of 49 stalks in half, would split them into two piles of 48 and 1 stalks respectively), but the data shows that a significant divergence between 7 and 8 (yin and yang) lines begins much earlier, at around x=15. This value of x describes a situation where a disparity of 49/15=3.26 stalks between divided piles would be tolerated, a level of ‘human error’ that is not at all unreasonable to expect (indeed, arguably the whole system depends on such human ‘noise’). Moreover, the probability of obtaining a 6 (Old Yin) line, the least probable outcome, actually doubles from 2 to 4 percent between x=20 and x=14, a very significant shift.

The obvious conclusion from the data is that x plays a non-trivial role in the distribution of probabilities when using the yarrow-stalk method of I Ching divination.

It is important to note that what has been demonstrated is not merely that the division of the stalks acts as the ‘random seed’ for the judgment: this much is obvious. The significant result has been the revealing of a continuous, systematic change in the distribution of probabilities correlated with a variable that could be a possible candidate for subconscious influence. For instance, a line of interpretation that could be developed is that a mind in turmoil is more conducive to a disparate split, so increasing the likelihood of obtaining the Old Yin and Young Yang lines.

The results suggest positioning the given probabilities for the I Ching system within a larger field of ‘possible systems’, with different values of x, and different values of P also. In this expanded context, the probabilities would represent the basic, continuously variable quantities of the system, with their distribution being limited by the choice of x and P. The divergences shown above would form a local feature of this expanded system.

(We might want to suggest a range of x that is to be considered ‘realistic’. One issue that needs to be addressed is the fact that as x increases, it becomes increasingly possible that one of the piles will fail to yield any line, because it contains too few stalks, which may skew the results. That said, however, the point of divergence at x=15 would surely fall within such a ‘realistic’ range.)

### The Hexagrams Re-Ordered

The real question is whether the divergences in the distribution of probabilities relate to any cogent change in the sorts of hexagrams, and therefore judgments, likely to result.

What remains then, is to process the probabilities of the hexagrams again, this time factoring in different values of x and if possible including moving lines. The results from those values of x corresponding to the significant divergence points of the data will then be examined to see the results of these divergences in the judgments.

### Part III

Full results (machine-processed, but programmed by me so it still might be wrong ;)) of the yarrow-stalk I Ching, ordered by probability (and secondarily by value).1 (Meanwhile please note that my original post had several errors, but the point still stands that the two ‘poles’ of K’un and Ch’ien are statistically separated by an order of magnitude when using the yarrow stalks, whereas using the coins they would have exactly the same probability of turning up. Hence I would suggest the use of coins has to be regarded as an degraded mode of access to the I Ching that gives the false impression of the absence of internal systemic bias or difference.) The asymmetrical probability of obtaining yin or yang lines has as its consequence a rather nice asymmetrical clustering of the probabilities of these binary values occurring:

```{63} 117649 /
hexagrams with only broken lines 16777216
{62,61,59,55,47,31} 84035 /
hexagrams with one whole line 16777216
{60,58,57,54,53,51,46,45,43,39,30,29,27,23,15} 60025 /
hexagrams with two whole lines 16777216
{56,52,50,49,44,42,41,38,37,35,28,26,25,22,21,19,14,13,11,7} 42875 /
hexagrams with three broken, three whole lines 16777216
{48,40,36,34,33,24,20,18,17,12,10,9,6,5,3} 30625 /
hexagrams with two broken lines 16777216
{32,16,8,4,2,1} 21875 /
hexagrams with one broken line 16777216
{0} 15625 /
hexagrams with only whole lines 16777216
```

The same clustering applied to the hexagram’s order numbers in the book:

```{2}
{24,7,15,16,8,23}
{19,36,46,51,40,62,3,29,39,45,27,4,52,35,20}
{11,54,55,32,60,63,48,17,47,51,41,22,18,21,64,56,42,59,53,12}
{34,5,58,49,28,26,38,30,50,61,37,57,25,28,33}
{43,14,9,10,13,44}
{1}
```
```Hexagram Name // Binary // Value // Order // Probability

___ ___
___ ___
___ ___
___ ___
___ ___
___ ___

K'un (The Receptive) 63 2 117649/16777216 (approx 1/142)

___ ___
___ ___
___ ___
___ ___
___ ___
_______
Fu (Returning) 62 24 84035/16777216 (approx 1/199)

___ ___
___ ___
___ ___
___ ___
_______
___ ___
Shih (The Army) 61 7 84035/16777216 (approx 1/199)

___ ___
___ ___
___ ___
_______
___ ___
___ ___
Ch'ien (Modesty) 59 15 84035/16777216 (approx 1/199)

___ ___
___ ___
_______
___ ___
___ ___
___ ___
Yu (Harmony) 55 16 84035/16777216 (approx 1/199)

___ ___
_______
___ ___
___ ___
___ ___
___ ___
Pi (Union) 47 8 84035/16777216 (approx 1/199)

_______
___ ___
___ ___
___ ___
___ ___
___ ___
Po (Falling Apart) 31 23 84035/16777216 (approx 1/199)

___ ___
___ ___
___ ___
___ ___
_______
_______
Lin (Approach) 60 19 60025/16777216 (approx 1/279)

___ ___
___ ___
___ ___
_______
___ ___
_______
Ming I (Darkening of the Light) 58 36 60025/16777216 (approx 1/279)

___ ___
___ ___
___ ___
_______
_______
___ ___
Sheng (Ascending) 57 46 60025/16777216 (approx 1/279)

___ ___
___ ___
_______
___ ___
___ ___
_______
Chen (Thunder) 54 51 60025/16777216 (approx 1/279)

___ ___
___ ___
_______
___ ___
_______
___ ___
Hsieh (Removing Obstacles) 53 40 60025/16777216 (approx 1/279)

___ ___
___ ___
_______
_______
___ ___
___ ___
Hsiao Kuo (Small Excesses) 51 62 60025/16777216 (approx 1/279)

___ ___
_______
___ ___
___ ___
___ ___
_______
Chun (Initial Difficulty) 46 3 60025/16777216 (approx 1/279)

___ ___
_______
___ ___
___ ___
_______
___ ___
K'an (The Perilous Pit) 45 29 60025/16777216 (approx 1/279)

___ ___
_______
___ ___
_______
___ ___
___ ___
Chien (Arresting Movement) 43 39 60025/16777216 (approx 1/279)

___ ___
_______
_______
___ ___
___ ___
___ ___
Ts'ui (Gathering Together) 39 45 60025/16777216 (approx 1/279)

_______
___ ___
___ ___
___ ___
___ ___
_______
I (Nourishment) 30 27 60025/16777216 (approx 1/279)

_______
___ ___
___ ___
___ ___
_______
___ ___
Meng (Youthful Inexperience) 29 4 60025/16777216 (approx 1/279)

_______
___ ___
___ ___
_______
___ ___
___ ___
Ken (Mountain) 27 52 60025/16777216 (approx 1/279)

_______
___ ___
_______
___ ___
___ ___
___ ___
Chin (Progress) 23 35 60025/16777216 (approx 1/279)

_______
_______
___ ___
___ ___
___ ___
___ ___
Kuan (Contempation) 15 20 60025/16777216 (approx 1/279)

___ ___
___ ___
___ ___
_______
_______
_______
T'ai (Peace) 56 11 42875/16777216 (approx 1/391)

___ ___
___ ___
_______
___ ___
_______
_______
Kuei Mei (The Marrying Maiden) 52 54 42875/1677721 (approx 1/391)

___ ___
___ ___
_______
_______
___ ___
_______
Feng (Abundance) 50 55 42875/16777216 (approx 1/391)

___ ___
___ ___
_______
_______
_______
___ ___
Heng (Perseverance) 49 32 42875/16777216 (approx 1/391)

___ ___
_______
___ ___
___ ___
_______
_______
Chieh (Regulation) 44 60 42875/16777216 (approx 1/391)

___ ___
_______
___ ___
_______
___ ___
_______
Chi Chi (Completion) 42 63 42875/16777216 (approx 1/391)

___ ___
_______
___ ___
_______
_______
___ ___
Ching (A Well) 41 48 42875/16777216 (approx 1/391)

___ ___
_______
_______
___ ___
___ ___
_______
Sui (Following) 38 17 42875/16777216 (approx 1/391)

___ ___
_______
_______
___ ___
_______
___ ___
K'un (Oppression) 37 47 42875/16777216 (approx 1/391)

___ ___
_______
_______
_______
___ ___
___ ___
Hsien (Influence) 35 51 42875/16777216 (approx 1/391)

_______
___ ___
___ ___
___ ___
_______
_______
Sun (Decrease) 28 41 42875/16777216 (approx 1/391)

_______
___ ___
___ ___
_______
___ ___
_______
Pi (Adornment) 26 22 42875/16777216 (approx 1/391)

_______
___ ___
___ ___
_______
_______
___ ___
Ku (Arresting Decay) 25 18 42875/16777216 (approx 1/391)

_______
___ ___
_______
___ ___
___ ___
_______
Shih Ho (Biting Through) 22 21 42875/16777216 (approx 1/391)

_______
___ ___
_______
___ ___
_______
___ ___
Wei Chi (Before Completion) 21 64 42875/16777216 (approx 1/391)

_______
___ ___
_______
_______
___ ___
___ ___
Lu (Travelling Stranger) 19 56 42875/16777216 (approx 1/391)

_______
_______
___ ___
___ ___
___ ___
_______
I (Increase) 14 42 42875/16777216 (approx 1/391)

_______
_______
___ ___
___ ___
_______
___ ___
Huan (Dispersion) 13 59 42875/16777216 (approx 1/391)

_______
_______
___ ___
_______
___ ___
___ ___
Chien (Gradual Progress) 11 53 42875/16777216 (approx 1/391)

_______
_______
_______
___ ___
___ ___
___ ___
P'i (Stagnation) 7 12 42875/16777216 (approx 1/391)

___ ___
___ ___
_______
_______
_______
_______
Ta Chuang (Power of the Great) 48 34 30625/16777216 (approx 1/547)

___ ___
_______
___ ___
_______
_______
_______
Hsu (Waiting) 40 5 30625/16777216 (approx 1/547)

___ ___
_______
_______
___ ___
_______
_______
Tui (Joy) 36 58 30625/16777216 (approx 1/547)

___ ___
_______
_______
_______
___ ___
_______
Ko (Revolution) 34 49 30625/16777216 (approx 1/547)

___ ___
_______
_______
_______
_______
___ ___
Ta Kuo (Excess) 33 28 30625/16777216 (approx 1/547)

_______
___ ___
___ ___
_______
_______
_______
Ta Ch'u (The Great Taming Force) 24 26 30625/16777216 (approx 1/547)

_______
___ ___
_______
___ ___
_______
_______
K'uei (Disunion) 20 38 30625/16777216 (approx 1/547)

_______
___ ___
_______
_______
___ ___
_______
Li (Clinging) 18 30 30625/16777216 (approx 1/547)

_______
___ ___
_______
_______
_______
___ ___
Ting (The Cauldron) 17 50 30625/16777216 (approx 1/547)

_______
_______
___ ___
___ ___
_______
_______
Chung Fu (Inmost Sincerity) 12 61 30625/16777216 (approx 1/547)

_______
_______
___ ___
_______
___ ___
_______
Chia Jen (The Family) 10 37 30625/16777216 (approx 1/547)

_______
_______
___ ___
_______
_______
___ ___
Sun (Gentle Penetration) 9 57 30625/16777216 (approx 1/547)

_______
_______
_______
___ ___
___ ___
_______
Wu Wang (Correctness) 6 25 30625/16777216 (approx 1/547)

_______
_______
_______
___ ___
_______
___ ___
Sung (Conflict) 5 28 30625/16777216 (approx 1/547)

_______
_______
_______
_______
___ ___
___ ___
Tun (Retreat) 3 33 30625/16777216 (approx 1/547)

___ ___
_______
_______
_______
_______
_______
Kuai (Removing Corruption) 32 43 21875/16777216 (approx 1/766)

_______
___ ___
_______
_______
_______
_______
Ta Yu (Great Possession) 16 14 21875/16777216 (approx 1/766)

_______
_______
___ ___
_______
_______
_______
Hsiao Ch'u (The Taming Force) 8 9 21875/16777216 (approx 1/766)

_______
_______
_______
___ ___
_______
_______
Lu (Treading Carefully) 4 10 21875/16777216 (approx 1/766)

_______
_______
_______
_______
___ ___
_______
T'ung Jen (Fellowship with Men) 2 13 21875/16777216 (approx 1/766)

_______
_______
_______
_______
_______
___ ___
Kou (Coming to Meet) 1 44 21875/16777216 (approx 1/766)

_______
_______
_______
_______
_______
_______
Ch'ien (The Creative) 0 1 15625/16777216 (approx 1/1073)
```
1. By ‘value’ I mean the hexagrams’ binary value as per Leibniz.