**A Direct Yarrow Stalk Method**

The traditional yarrow stalk method involves dividing a bundle of 49 stalks, then counting through both heaps of stalks while carefully saving the remainders. However, most of this process is simply ritual, a kind of homage to numerology, because for each step, the outcome is determined as soon as the bundle is divided. Some prefer the traditional method, saying that the ritual helps maintain the proper mindset. However, here is an alternative that goes directly to the result of the division of the bundle, skipping the noncontributory steps while preserving the feel of the traditional method. It requires half as many counting off steps, the math is simpler, and the results are identical.

• Start with a bundle of approximately 50 stalks.

• Divide the bundle into two halves and set one of the halves down.

• Count off the other half by fours until 4 or fewer stalks remain.

• Set the remainder down in a pile by itself.

• Pick up the rest of the stalks and repeat the above process twice.

• You now have three remainder piles of 1 to 4 stalks each.

• Remove one stalk from the first remainder pile, unless there is only one.

• For each pile, assign a value of 3 of it contains 1 or 2 stalks, and 2 if it contains 3 or 4.

• The values add up to 6, 7, 8, or 9 for changing yin, yang, yin, or changing yang, respectively.

• Repeat the entire process five times to complete the hexagram.

**Notes**

The exact number of stalks is not important, as long as there are enough to divide randomly and count off by fours.

For results identical to the traditional method, count off the half that would not have one stalk removed at the beginning, which is the left half according to most instructions.

Removing one stalk from the first remainder pile makes all the value assignments the same. Otherwise, the first pile has a value of 3 if it contains 1 to 3 stalks, and 2 if it contains 4. The extra 3 is what skews the yarrow stalks toward changing yang. If you skip this step and treat all the piles the same, the odds match the 3-coin method.

For the value assignments, think: small pile, big value; big pile, small value. (Just like in the traditional method.)

The only difference between this and the traditional method are ritual steps which do not affect the outcome. The results are identical.

**Yarrow stalk mathematics**

The sizes of the remainder piles, on which the results of yarrow stalk consultation are based, are very consistent. After the first division, the remainder pile will have 5 (small) or 9 (large) stalks; after all other divisions, the pile will have 4 (small) or 8 (large) stalks. Here is how they add up. It is assumed that one stalk is set aside from the right heap.

First division: | |||||

one stalk set aside | 1 | 1 | 1 | 1 | |

right heap remainder | 1 | 2 | 3 | 4 | |

left heap remainder | 3 | 2 |
1 | 4 | |

total | 5 | 5 | 5 | 9 | |

All other divisions: | |||||

one stalk set aside | 1 | 1 | 1 | 1 | |

right heap remainder | 1 | 2 | 3 | 4 | |

left heap remainder | 2 | 1 |
4 | 3 | |

total | 4 | 4 | 8 | 8 | |

Remainder values: | |||||

small pile (4 or 5) | 3 | ||||

large pile (8 or 9) | 2 |

Note that the left heap remainder has a size correspondence with the total, and can be used as a substitute for the entire process. After the first division, 1, 2, or 3 stalks constitute a small remainder pile, and 4 constitute a large. After the other divisions, 1 or 2 are small, and 3 or 4 are large.

This is the basis of the direct method above. The only effect that counting off both heaps has is to separate the sizes of the large and small remainder piles.

The exact number of stalks is only important if both heaps are counted off, because then the remainders of both heaps must correspond to form the totals above.

If you use the direct method above with the right heap without first setting aside one stalk, it still works, and the odds are equivalent; but the actual outcome of the stalk division is changed, because the remainders from the divisions in the tables above change from 1, 2, 3, 4 to 2, 3, 4, 1.

**The minimalist method**

• Start with a pile of 25-50 stalks.

• Divide the pile (or even simpler, just pick up a convenient handful) and count them off by twos until there are just two or three left.

• Repeat twice to make a total of 6, 7, 8, or 9 remainder stalks.

This is how I use yarrow stalks. Works with any convenient objects such as toothpicks. Hua-Ching Ni describes a similar method using small seeds.

The odds match the three-coin method. To preserve the traditional yarrow stalk odds, count off the first handful by fours until there are 2, 3, 4, or 5 left; then if there are 4 or 5, remove 1 or 2 to make 3.

One last note: the “traditional” yarrow stalk method is not necessarily the “original” one.

**Pure conjecture on why this could be the original method**

Hypothesis: the three-coin method was invented in ancient times to duplicate the original yarrow stalk method, and retains its essential features. So why were the numbers 2 and 3 used to form the lines, when 0 and 1 work just as well? Before coins came into everyday use, yarrow stalks or other such objects could be used to make a random choice between two “heads or tails” options. One simply chose a random number of stalks and determined whether the number was odd or even. Counting the stalks off by twos until there were either 0 or 1 left would work; but a remainder pile of zero stalks does not actually constitute a physical representation of the process. So if the number is even, we stop short of 0, and leave a remainder of 2. But now the even remainder 2 (yin) exceeds the odd remainder 1 (yang), which (in a patriarchal society, at least) is a reversal of the accepted order. So, if the number is odd, we stop short of 1, and leave a remainder of 3. These numbers are conveniently small enough to be instantly recognizable as odd or even, and thus serve their purpose. When coins came along, each side of a coin was assigned either a 2 or a 3 in rote mimicry of the yarrow stalk method, and we still do it today.

*I Ching* “bibliomancy”

• Fan the pages of a book (such as the *I Ching*).

• Very casually divide the pages at a random position, as if cutting a deck of cards.

• Look at the last two digits of the number of the even-numbered page (generally on the left).

• If it is divisible by four, count it as “tails” or 2.

• If not, count it as “heads” or 3.

• To make the math easier, you can subtract a multiple of 4 from the page number, such as 40 or 80.

• Three divisions determine one line, just like using coins.

Note: this isn’t actually bibliomancy; it is sortilege using the pages of a book.

The edges of the pages must be fairly clean and smooth, without irregularities that could throw off the results.

**Yet another method, which requires counting-off only once per line**

• Start with a bundle of approximately 50 stalks.

• Divide the bundle into thirds, and set one of the thirds down in its own pile.

• Combine the other two thirds, then divide them into thirds again. Set one of those thirds down with the original third. This combination is the bundle that will be used from this point on; discard the rest.

• Separate six stalks from the bundle and set them aside in their own pile.

• Count off and discard the rest in a repeating pattern of 2, 2, 4 until there is at least one stalk left in your hand. You will be left with 1, 2, 3, or 4 stalks in addition to the 6 set aside at the beginning.

• If there are 4 stalks left in your hand (the next number to be counted off being 4), discard them all.

• You are now left with 0, 1, 2, or 3 stalks in your hand. These plus the six set aside at the beginning total 6, 7, 8, or 9. The odds match the 3-coin method.

**Notes**

This method involves counting off in a repeating pattern which totals 8 stalks, so two divisions are performed at the beginning to make the result a little more random. Don’t be precise with the divisions; just let the stalks fall apart by themselves.

The possibilities for stalks remaining in your hand, depending on the last number to be counted off, are as follows:

2: 1, 2

2: 1, 2

4: 1, 2, 3, 4

These, plus the six set aside at the beginning, total 7, 8, 9, or 10, with triple the possibilities for 7 or 8. We just need to convert the 10 to a 6 by discarding a remainder of 4 to turn the results into 6, 7, 8, or 9.

You could vary the result by counting in a pattern of 2, 4, 2 or 4, 2, 2 depending on how you feel at the moment.