The Markov chain of readings
Where does a series of readings drift?
Press Simulate to chain readings and see which stationary distribution each method tends to.
If we chain consultations (taking the future hexagram of one as the present of the next) we have a Markov chain over the 64 states. Each line mutates according to its method (derived from coins against yarrow). The question: does the asymmetry of the yarrow bias where the series travels? The stationary distribution (where it tends in the long run) answers it, and the Perron-Frobenius theorem guarantees it is unique: since the transition matrix is positive, it has a simple, dominant eigenvalue 1 (the second modulus is exactly 0.5, the same mixing speed in both methods).
all sizes equal: uniform stationary distribution (1/64)
Kun (yin) dominates; Qian (yang) almost vanishes: a bias to yin
| Method | π(Kun) | π(Qian) | Kun / Qian | expected yang | corr(π, yang) | λ₂ |
|---|---|---|---|---|---|---|
| Three coins | 1.56% | 1.563% | 1 | 3.00 | 0.00 | 0.50 |
| Yarrow (49 stalks) | 17.80% | 0.024% | 729 | 1.50 | −0.73 | 0.50 |
With coins the stationary distribution is uniform (1/64 per hexagram, correlation 0). With yarrow it is not: it skews towards yin, with correlation −0.73 between probability and number of yang lines. Kun is 729 times more likely than Qian, and the expected number of yang falls from 3.0 to 1.5. The cause is that old yang (3/16) mutates more than old yin (1/16): yang decays towards yin. Curiously both methods mix at the same speed (λ₂ = 0.5, relaxation 2 steps): equally fast, but to different destinations.
It starts at Qian (6 yang) and chains 200,000 consultations: at how many yang lines does the average settle?
The Markov chain of consultations has a uniform stationary distribution with coins and one skewed to yin with yarrow (Kun is 729 = 3^6 times more likely than Qian); both mix at the same speed, with a simple eigenvalue 1 and second modulus 0.5.
verificar_markovThe types of claim and the full bibliography (APA) are in Foundations.