Coin Oracle

General Procedure

As I mentioned earlier, there are generally two methods for consulting the I Ching. One of them is the coin oracle. This oracle is performed using three coins. There is a more traditional, and hence preferred, oracle called the stalk oracle, but I have chosen to analyze the coin oracle first because the analysis is a bit simpler than it is for the stalk oracle. The stalk oracle is analyzed in detail in the next chapter.

Traditionally, the coins used for the oracle are bronze with a hole in the middle. There is an inscription on one side of each coin.

All three coins are tossed once to generate a single line. Thus, six tosses of the coins are required to generate all six lines of a single hexagram.

The lay of the coins after each toss is interpreted in the following way. Any coin which lands with the inscription up is interpreted as yin, and thus has a value of two (2). Any coin which lands with the inscription down is interpreted as yang, and thus has a value of three (3). The reasons for the values of 2 and 3 have origins in numerology. I must refer you to the references for discussion of those values. It will suffice to say here that all the even numbers are yin numbers and all the odd numbers are yang numbers. Anyway, the actual values used to represent the lines do not affect the probability analysis at all.

The table below indicates all of the possible combinations of a coin toss and the resulting value.

yin + yin + yin
yin + yin + yang
yin + yang + yin
yin + yang + yang
yang + yin + yin
yang + yin + yang
yang + yang + yin
yang + yang + yang

There are exactly eight possible combinations for each toss of the three coins.

The evaluation of the toss is done by adding the values of all the coins to determine the line associated with it. For example, the above combinations translate into the following sums:

2 + 2 + 2 = 6
2 + 2 + 3 = 7
2 + 3 + 2 = 7
2 + 3 + 3 = 8
3 + 2 + 2 = 7
3 + 2 + 3 = 8
3 + 3 + 2 = 8
3 + 3 + 3 = 9

You may note that some of these combinations result in identical sums. For example, yin + yin + yang (2 + 2 + 3) is the same as yin + yang + yin (2 + 3 + 2) since it consists of two yin states and one yang state. We end up with four possible summations out of the eight possible coin configurations. In the table below I have noted how many of the eight combinations will result in each sum.

2 + 2 + 2 = 6 one combination
2 + 2 + 3 = 7 three combinations
3 + 3 + 2 = 8 three combinations
3 + 3 + 3 = 9 one combination

The resulting numbers come from the set {6, 7, 8, 9} with varying probabilities. These numbers are called "Ritual Numbers". They have meaning unto themselves, however that discussion is out of the scope of this treatise. I refer you to the references. The line itself is determined by mapping the above set of numbers into the lines using the following table:

6 moving yielding line
7 static firm line
8 static yielding line
9 moving firm line

You may notice that both of the even numbers (6 and 8) map to the yielding (yin) lines and that both of the odd numbers (7 and 9) map to the firm (yang) lines.

So, the entire hexagram is built from the bottom up by tossing the coins and calculating the sum which indicates one of the four types of lines.

The following probability analysis will answer some interesting questions about the frequency of occurrences of hexagrams with and without moving lines.

Probability Analysis

I became curious about the probability of drawing each type of line in the hexagram, the likelihood of drawing static hexagrams or hexagrams with moving lines, the probabilities of drawing some unique hexagrams, and other situations. This section shows how to calculate those probabilities for the coin oracle.

If you look at the tables above, you can see that out of 8 outcomes, the hexagram line numbers occur with the following frequencies:

6 is 1 in 8
7 is 3 in 8
8 is 3 in 8
9 is 1 in 8

If we assume that the probability for a yin or yang outcome for any single coin toss is equal (which it is if the coins are not loaded), we can see that a toss of the three coins is three times as likely to produce a static line as a moving line (the 6 and the 9 are moving lines).

We can also see that it is equally probable to obtain a yielding line or a firm line. Since the 6 and the 8 both represent a yielding line, we have 4 chances in 8 of obtaining a yielding line. Similarly, since 7 and 9 both represent a firm line, we have 4 chances in 8 of obtaining a firm line. The probability of yielding lines versus firm lines is equal, however, the probability of moving versus static lines is not.

In combinatoric calculations, the probability of any specific outcome is usually expressed as a decimal value less than 1.0. The value represents the percentage of times, on average, that a specific outcome will occur. So, the probability of obtaining a yin value from the toss of a single coin is 0.5, or 50%, or one half. The probability of obtaining a yang value for a toss of the coin is also 0.5. If all of the probabilities of all the outcomes are added together, they must add up to exactly 1.0. You can see that 0.5 and 0.5 add up to 1.0. The probability value of 1.0 is certainty. This means that we are certain to toss either a yin or a yang since the coin has only two sides (we discount the possibility of the coin standing on edge, or at least we would reaccomplish the toss in that situation).

Using this means of expression, we can observe that the following probabilities are true.

Probability of a yielding line = 0.5 (one half, or 4 of 8)
Probability of a firm line = 0.5 (one half, or 4 of 8)
Probability of a moving line = 0.25 (1 in 8 plus 1 in 8 = 2 in 8)
Probability of a static line = 0.75 (3 in 8 plus 3 in 8 = 6 in 8)
Probability of a moving yin line = 0.125 (just the moving yin, 1 of 8)
Probability of a moving yang line = 0.125 (just the moving yang, 1 of 8)

These probabilities apply to a single line. Using combinatoric mathematics, we can calculate the probability of any combination of any number of lines. Let us address the case for two lines.

With two lines, we must apply some simple math to the probabilities which we know for the single lines. For a single line, the combinations of coins are:

2 + 2 + 2 = 6 moving = M1
2 + 2 + 3 = 7 static = S1
2 + 3 + 2 = 7 static = S2
2 + 3 + 3 = 8 static = S3
3 + 2 + 2 = 7 static = S4
3 + 2 + 3 = 8 static = S5
3 + 3 + 2 = 8 static = S6
3 + 3 + 3 = 9 moving = M2

Notice that 2 of the 8 are moving lines and 6 of the 8 are static lines. I have given names to each of the outcomes so I can make a table of them. Because we use the same procedure to determine the second line as we do for the first, the draw of the second line also has 8 possible outcomes. They are the same outcomes as for the first line. The table of outcomes is:

M1 S1 S2 S3 S4 S5 S6 M2
M1 * *
M2 * *

Each entry in the table represents drawing the first line and obtaining the indicated outcome in the vertical column and then subsequently drawing the second line and obtaining the indicated outcome in the horizontal row. Thus two lines are obtained, and the table lists all of the possible outcomes for the two draws. There are 64 entries in the table. I have marked the entries in the table in which two moving lines have been obtained sequentially by using a *. There are four of them. Therefore, out of 64 possible outcomes, only 4 of them result in both lines moving. Thus, the probability of drawing two moving lines in a row is 4/64, or 1/16, or 0.0625.

There is an easier way to calculate this result. As it turns out, the same mathematical result will be obtained if we simply multiply the probability of drawing a single moving line times itself for each time we draw another line and require it to be a moving line. Therefore, if we multiply the probability of drawing one moving line (0.25) times the probability of drawing another moving line (0.25) we obtain the same result (0.0625). This procedure can be generalized to any number of moving lines, for example the probability of drawing an entire trigram (three lines) of only moving lines is:

0.25 x 0.25 x 0.25 = 0.015625

The procedure can be generalized even further. Suppose we want to know the probability of drawing exactly one moving line and then drawing exactly one static line. We simply multiply the two probabilities together. For example, the probability of drawing a moving line is 0.25, and the probability of drawing a static line is 0.75. Therefore, the probability of drawing a moving line and then a static line is:

0.25 x 0.75 = 0.1875

If you consult the table of outcomes above, you will see that there are exactly 12 entries which represent drawing a moving line and then subsequently drawing a static line. So, the probability is 12/64 = 0.1875.

The probability of drawing the converse outcome, one static line followed by one moving line is:

0.75 x 0.25 = 0.1875

This is the same value as before, but it represents a different sequence of events. It is clear then that drawing a moving line followed by a static line is equally probable as drawing a static line followed by a moving line.

Let us suppose that we will accept either order of the lines, moving then static or static then moving, and we want to know what the probability is of that outcome. We simply add the two probabilities together. So, the probability of drawing either moving then static or static then moving is:

0.1875 + 0.1875 = 0.375

This is equivalent to adding up all the entries in the table of outcomes which represent those specific events.

So, it appears that we now can calculate the probability of drawing any combination of two lines with either or both of the lines moving. Here are the outcomes which represent that situation:

moving then static
static then moving
moving then moving

The only combination left out is static then static, which contains no moving lines. We will get to that in a minute.

Right now, we can calculate the probability of obtaining any moving lines at all in the two lines. We simply add the three probabilities above together to get:

0.1875 + 0.1875 + 0.0625 = 0.4375

At this point we have the probability of drawing two lines of a hexagram and having either one or both be moving lines. Let's calculate the converse outcome, namely that neither of the lines is a moving line. This is surprisingly simple. Simply multiply together the probability of drawing a static line and then drawing another static line:

0.75 x 0.75 = 0.5625

So now we have two probability figures. One (0.4375) is for the case where one or more lines are moving, and the other (0.5625) is the case where no lines are moving. Since this covers all possible outcomes of the draws (you either have moving lines or you don't), the probabilities should add up to 1.0:

0.5625 + 0.4375 = 1.0

Since 1.0 is certainty, we have shown that, mathematically speaking, we will certainly draw one of the two situations calculated above. Either some of the two lines move, or none of them do.

So now we can answer the question, "Which is more likely? All static lines, or some moving lines?". The answer is that drawing two lines which are both static is slightly more likely than drawing two lines and having some of them move. This is true when using the coin oracle.

Since these two outcomes are mutually exclusive (either we have moving lines or we do not), there is an even simpler way of calculating some of the numbers. For example, we can find out that drawing two static lines in a row has the probability of

0.75 x 0.75 = 0.5625

and since we are certain to draw either some moving lines or no moving lines, we can calculate the other figure (some moving lines) by simply subtracting our first calculations from certainty:

1.0 - 0.5625 = 0.4375

Now that the foundation has been laid, we can calculate some interesting probabilities. For example, one unique hexagram is:

This is hexagram 1, the Creative, with all the lines moving. It transitions into hexagram 2, the Receptive. This is a fairly propitious and significant hexagram to draw. So we can use the technique described above to calculate the probability of this hexagram. The probability of drawing a moving yang line is 0.125. So the probability of drawing 6 of them in a row is:

0.125 x 0.125 x 0.125 x 0.125 x 0.125 x 0.125 = 0.000003814697265625

which is not very likely at all. This value is about 3.81e-6, which is about one in 262,144.

This calculation is an interesting result. Since the hexagram above is just one of 4096 possible outcomes, one might think that the probability of obtaining it would be one in 4096 or 0.000244140625, which is considerably more likely than the calculation above. However, this is not the case. All possible arrangements of all the hexagrams do not have equal probabilities when the moving lines are taken into account. This is because you are less likely to draw a moving line than a static line. That is the way the coin oracle operates. For example, let us calculate the probability of drawing hexagram 1 with no moving lines at all.

The probability of drawing one static yang line is 3/8, or 0.375. Hence, the probability of drawing 6 static yang lines in a row is:

0.375 x 0.375 x 0.375 x 0.375 x 0.375 x 0.375 = 0.002780914306641

This is about 2.78e-3, or one in about 360. As you can see, this is a higher probability than a simple 1 in 4096 calculation. It is also higher than the probability of drawing hexagram 1 with all moving lines. We can conclude that drawing hexagram 1 with no moving lines is about 729 times as likely as drawing hexagram 1 with all moving lines.

What about the probability of drawing hexagram 1 with no moving lines as opposed to drawing that same hexagram with one or more moving lines?

We can calculate this by noting that drawing hexagram 1 without any consideration about whether or not the lines move is 1 in 64. I could demonstrate this fact by calculating the probability for each of the 64 combinations of moving and static lines for the hexagram and adding them together. However, to save space I will forgo that calculation.

This probability of obtaining any one of the 64 hexagrams when no account is given to the motion or stasis of the lines is 1/64 or 0.015625. All hexagrams are equally likely if moving lines are not taken into account. Now, if we disallow the single unique case wherein the hexagram has no moving lines at all, we can subtract the probability of 6 static yang lines in a row from the 1/64 value and get:

0.015625 - 0.002780914306641 = 0.01284408569336

This is about one in 78. Hence, it is much more likely that if we draw hexagram 1 it will have at least some moving lines than no moving lines at all. Indeed, it is about 4.62 times as likely to draw the hexagram with some moving lines as it is to draw the same hexagram with no moving lines.

Here are the probabilities we have calculated

Probability of any hexagram with some moving lines is 1 in 78
Probability of any hexagram with no moving lines is 1 in 360
Probability of any hexagram with all moving lines is 1 in 262,144

This conclusion can actually be applied to any individual hexagram since from a probability perspective, any of the 64 hexagrams is equally likely to occur if the movement of the lines is discounted. There is nothing mathematically unique about hexagram 1 when dealing with the coin oracle.

Therefore, the conclusions we can draw from this calculation are that drawing any specific hexagram with no moving lines at all is about 729 times as likely as drawing that same hexagram with all moving lines. However, drawing the same hexagram with some moving lines is about 4.6 times as likely as drawing the hexagram with no moving lines at all.

It therefore seems that for any specific hexagram, the most likely outcome with the coin oracle is to draw that hexagram with some moving lines in it.

I will now analyze the stalk oracle and compare it to the coin oracle.

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