wandelaar Posted July 5, 2018 (edited) In this topic I want to calculate the theoretical probabilities of the hexagrams when the yarrow stalk method is used. Now the manner of division of the heap of stalks will differ slightly from person to person: some people will strife to divide the heap nearly in half while others will be more nonchalant about it. That means that the probabilities for the possible divisions will also differ slightly from person to person, and as a further consequence the probabilities of obtaining the different line-types will differ slightly from person to person. I have considered different ways to tackle this problem, but have decided that the simplest approach is taking the probabilities of obtaining the different line-types as free parameters in the calculation. When someone wants to use the formulae of this topic he will have to plug in the probabilities of obtaining the different line-types as they are when that person throws the I Ching by the yarrow stalk method without actual consultation. Determining those personal probabilities will then be an empirical matter. Edited July 5, 2018 by wandelaar 1 Share this post Link to post Share on other sites

wandelaar Posted July 5, 2018 (edited) We will use the following symbols: p6 = probability of throwing a moving yin line p7 = probability of throwing a static yang line p8 = probability of throwing a static yin line p9 = probability of throwing a moving yang line The numbers 6, 7, 8 and 9 are chosen so as to correspond to the usual numerical representations of the line-types in the I Ching. The actual probabilities depend on the person throwing the I Ching and have to be empirically estimated by letting the person throw the I Ching a large number of times (but without posing questions or consulting the I Ching) and calculating the relative frequencies of the line-types. Edited July 5, 2018 by wandelaar 1 Share this post Link to post Share on other sites

wandelaar Posted July 5, 2018 (edited) We can represent all hexagrams by means of a row vector H = (f,e,d,c,b,a) where the numbers a, b, c, d, e and f can take on all of the values 6, 7, 8 and 9 representing the different line-types in the usual manner. The number a represents the lowest line in the hexagram and the number f represents the highest line of the hexagram. Example: Edited July 5, 2018 by wandelaar 1 Share this post Link to post Share on other sites

Lost in Translation Posted July 5, 2018 The probability of obtaining a hexagram is the product of the probabilities of obtaining its individual lines. For example: 1 hour ago, wandelaar said: We can represent all hexagrams by means of a row vector H = (f,e,d,c,b,a) where the numbers a, b, c, d, e and f can take on all of the values 6, 7, 8 and 9 representing the different line-types in the usual manner. The number a represents the lowest line in the hexagram and the number f represents the highest line of the hexagram. For example, you list (f,e,d,c,b,a) = (8,9,7,6,7,8). Previously we showed various ways to calculate the likelihood of a given line being received. Taking the standard textbook percentages, e.g. Quote 3/16 = 18.75% (9: old yang) 5/16 = 31.25% (7: young yang) 1/16 = 6.25% (6: old yin) 7/16 = 43.75% (8: young yin) we can reformulate this as (f,e,d,c,b,a) = (43.75%, 18.75%, 31.25%, 6.25%, 31.25%, 43.75%), which yields .000219047 or .0219047 %. You will quickly notice that the order of the lines has no relevance in this calculation, so hexagrams composed of the same lines but in different order will yield the same percentages. I do not know how to account for this factor. Share this post Link to post Share on other sites

wandelaar Posted July 5, 2018 (edited) A hexagram is build up from the lowest line upwards. Now when somebody throws hexagrams (without consultation!) a great many times (let's say N times) we may expect that the relative frequencies of the hexagrams found approximate the probabilities of those hexagrams on the basis of pure chance. Those probabilities are the person-specific numbers p6, p7, p8 and p9 . Let's look at the specific hexagram represented by (f,e,d,c,b,a) = (8,9,7,6,7,8). For this hexagram to turn up the following has to happen: 1. Line 1 has to be of the type represented by 8. This happens with an approximate relative frequency of p8 . 2. Line 2 has to be of the type represented by 7. This happens with an approximate relative frequency of p7 . 3. Line 3 has to be of the type represented by 6. This happens with an approximate relative frequency of p6 . 4. Line 4 has to be of the type represented by 7. This happens with an approximate relative frequency of p7 . 5. Line 5 has to be of the type represented by 9. This happens with an approximate relative frequency of p9 . 6. Line 6 has to be of the type represented by 8. This happens with an approximate relative frequency of p8 . So we expect line 1 of a thrown hexagram to be correct approximately p8 * N times; we expect both line 1 and line 2 of a thrown hexagram to be correct approximately p7 * (p8 * N) times; we expect all of the lines 1, 2 and 3 of the hexagram to be correct approximately p6 * ( p7 * (p8 * N)) times; and so on. Thus we expect approximately { p8 * p9 * p7 * p6 * p7 * p8 } * N of the thrown N hexagrams to be the hexagram represented by (f,e,d,c,b,a) = (8,9,7,6,7,8). The relative frequency of the hexagram of our example is thus expected to approximate: [{ p8 * p9 * p7 * p6 * p7 * p8 } * N]/N = p8 * p9 * p7 * p6 * p7 * p8 . And this expected approximation can be made as precise an one wants it by taking N large enough. So the probability of finding the hexagram represented by (f,e,d,c,b,a) = (8,9,7,6,7,8) is p8 * p9 * p7 * p6 * p7 * p8 . Generalising the above reasoning we get the following simple formula for the probability P(f,e,d,c,b,a) of throwing the hexagram represented by (f,e,d,c,b,a) : P(f,e,d,c,b,a) = pf * pe * pd * pc * pb * pa (This is the product referred to by Lost in Translation in his previous post.) Edited July 5, 2018 by wandelaar 1 Share this post Link to post Share on other sites

wandelaar Posted July 5, 2018 36 minutes ago, Lost in Translation said: I do not know how to account for this factor. What factor do you mean? Share this post Link to post Share on other sites

Lost in Translation Posted July 5, 2018 28 minutes ago, wandelaar said: What factor do you mean? How to account for line order. E.G. 678987 has same probability as 898677 or 789876 or 789687 etc. since they have the same component lines. Share this post Link to post Share on other sites

wandelaar Posted July 5, 2018 31 minutes ago, Lost in Translation said: How to account for line order. E.G. 678987 has same probability as 898677 or 789876 or 789687 etc. since they have the same component lines. Yes - that has to be so, because the lines are thrown independently. Why should that be a problem? Share this post Link to post Share on other sites

Lost in Translation Posted July 5, 2018 1 hour ago, wandelaar said: 2 hours ago, Lost in Translation said: How to account for line order. E.G. 678987 has same probability as 898677 or 789876 or 789687 etc. since they have the same component lines. Yes - that has to be so, because the lines are thrown independently. Why should that be a problem? It's not a problem. I am simply stating that it is so. 1 Share this post Link to post Share on other sites