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)