The Probabilities of the 4096 Changes:
Coin Oracle

We have seen that the coin probabilities of the four hyperlines are:
6 -x- old yin 1/8
7 --- young yang 3/8
8 - - young yin 3/8
9 -o- old yang 1/8

After casting a hyperhexagram (a change) with six tosses of three coins, we may record the result as a hexcode. For example, "678969" means a six (old or changing yin) in the bottom line, a seven (young or unchanging yang) in the second line, and so on, with a nine (old or changing yang) on the top.

If this change is obtained by choosing two hexagrams at random, then its probability is 1/4096, or 0.000244140625, as all 4096 changes are equally likely.

But if the same change is obtained by six tosses of three coins, the probability is different, as the "7" and "8" are (as we have seen) three times as likely as the "6" or the "9". We may calculate the probability of this change from its hexcode, "678969", by multiplying the probability of the hyperline codes in order,

p(6) * p(7) * p(8) * p(9) * p(6) * p(9) =
1/8 * 3/8 * 3/8 * 1/8 * 1/8 * 1/8 =
1 * 3 * 3 * 1 * 1 * 1 / 8 ^ 6 =
9 / 2 ^ 18 = 9 / (64 * 4096) =
9 / 262,144
where p(6) denotes the probability of obtaining a six. In other words, using coins this change is (9/64) times as likely as opening the book twice at random. We will refer to this factor, (9/64), as the "coin factor" of the change. Thus the coin probability of a change is the coin factor times the random probability.

From this example we may derive a general rule for the coin factor of a change: let n be the number of unchanging lines. Then 3 ^ n / 64 is the coin factor.

Note: The coin factor ranges from a minimum (in the case of all lines being changing) of 1/64 (corresponding to a very small probability of about 4 times in a million) to a maximum coin factor (in the case of all six lines being unchanging) of

3^6 / 64 = 729 / 64 = 11.390625
corresponding to a probability of about 3 times in a thousand.
Revised 18 Dec 2007 by Ralph