irrational numbers

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))&lt;a&lt;(.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))&lt;a&lt;(.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) =&gt; 9
16 (8+4+4) = (2+3+3) =&gt; 8
20 (8+8+4) = (2+2+3) =&gt; 7
24 (8+8+8) = (2+2+2) =&gt; 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 persumably 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)&lt;a&lt;(.5P+P/x),
r1 = (1 + a mod 4 + (b-1) mod 4),
a'+b'=(49-r1) and
(.5(P-r1))-(P-r1/x))&lt;(P-r1)&lt;(.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))&lt;(P-r1)&lt;(.5(P-r1-r2)+(P-r1-r2/x)),
r3 = (1 + a'' mod 4 + (b''-1) mod 4 );

<center> * * *</center>

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.

<img alt="graph.gif" src="http://blog.urbanomic.com/dread/archives/graph.gif" width="394" height="120" />

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.