Unfortunately, the method that was used in such ancient times has been lost. The procedure we use today is a reconstruction dating back the 12th century CE and is described in the commentaries that form the Ten Wings.

This method is quite laborious and requires some dexterity (to hold the yarrow stalks) and focus (to properly count them). This is both its strong and weak point: some find it too complex while others consider the time required as well spent meditating on the question.

The process starts with 50 yarrow stalks of which one is put aside and no longer used.

To get a line one has to proceed as follows:

- Split the 49 stalks in two groups;
- Take one stalk from the left group and put it aside;
- Count the left group by four until you have four or less stalks left in the group;
- Put the (one to four) remaining stalk together with the one you took on step 2;
- Count the right group by four until you have four or less stalks left in the group;
- Put the (one to four) remaining stalk together with the ones you got from step 2 and 5;
- If you remained with
*nine*stalks, mark 2, if you remained with*five*stalks mark 3; - Put the stalks you counted all together (they should be 40 or 44), split them in two groups and repeat steps 2-6;
- If you remained with
*eight*stalks, mark 2, if you remained with*four*stalks mark 3; - Put the stalks you counted all together, (they should be 32, 36 or 40) split them in two groups and repeat steps 2-6;
- If you remained with
*eight*stalks, mark 2, if you remained with*four*stalks mark 3; - Sum up the three numbers you got, the sum should be either
**6**,**7**,**8**or**9**, and draw the line according the following table.

6 7 8 9

An alternative method of counting is to ignore steps 7, 9 and 11 and group all the stalks you get in a single heap. After you have performed the split three times, you divide the stalks in the heap (which will contain eather 24, 28, 32 or 36 stalks) by four to directly get the number of the resulting line:

**6**,

**7**,

**8**,

**9**.

Actually this can be simplified further to avoid counting both groups. After step 3:

- if you get
*four*stalks, pick*four*stalks from the other group and count 2 - if you get
*three*stalks, pick*one*stalks from the other group and count 3 - if you get
*two*stalks, pick*two*stalks from the other group and count 3 - if you get
*one*stalks, pick*three*stalks from the other group and count 3

After step 8 (and 10):

- if you get
*four*stalks pick*three*stalks from the other group and count 2 - if you get
*three*stalks pick*four*stalks from the other group and count 2 - if you get
*one*stalks pick*two*stalks from the other group and count 3 - if you get
*two*stalks pick*one*stalks from the other group and count 3

Searching on YouTube will provide you with a great deal of video example on how to use the Yarrow stalks to get hexagram lines.

#### Probabilities

The probabilities for this method are known to be:
Prob(6) =

Prob(8) =

Prob(7) =

Prob(9) =

Prob(

^{1}/_{16}Prob(8) =

^{7}/_{16}Prob(7) =

^{5}/_{16}Prob(9) =

^{3}/_{16}Prob(

*yin*) = Prob(*yang*) =^{1}/_{2}on the basis of the following reasoning:

- On the first subdivision, 49 stalks, we can get 2 with a probability of
^{1}/_{4}and 3 with a probability of^{3}/_{4} - On the second and third subdivision, we can get 2 with a probability of
^{2}/_{4}and 3 with a probability of^{2}/_{4} - Hence the probabilties for each possible outcome are:
Prob(2+2+2) = ^{1}/_{4}*^{1}/_{2}*^{1}/_{2}=^{1}/_{16}Prob(2+2+3) = ^{1}/_{4}*^{1}/_{2}*^{1}/_{2}=^{1}/_{16}Prob(2+3+2) = ^{1}/_{4}*^{1}/_{2}*^{1}/_{2}=^{1}/_{16}Prob(3+2+2) = ^{3}/_{4}*^{1}/_{2}*^{1}/_{2}=^{3}/_{16}Prob(2+3+3) = ^{1}/_{4}*^{1}/_{2}*^{1}/_{2}=^{1}/_{16}Prob(3+2+3) = ^{3}/_{4}*^{1}/_{2}*^{1}/_{2}=^{3}/_{16}Prob(3+3+2) = ^{3}/_{4}*^{1}/_{2}*^{1}/_{2}=^{3}/_{16}Prob(3+3+3) = ^{3}/_{4}*^{1}/_{2}*^{1}/_{2}=^{3}/_{16} - Summing up the probabilities for each possible result, we get:
Prob(6) = Prob(2+2+2) = ^{1}/_{16}Prob(8) = Prob(2+3+3) + Prob(3+3+2) + Prob(3+2+3) = ^{1}/_{16}+^{3}/_{16}+^{3}/_{16}= ^{7}/_{16}Prob(7) = Prob(2+2+3) + Prob(2+3+2) + Prob(3+2+2) = ^{1}/_{16}+^{1}/_{16}+^{3}/_{16}= ^{5}/_{16}Prob(9) = Prob(3+3+3) = ^{3}/_{16}

Unfortunately, the analysis above is not accurate as it assumes that, for each subdivision, the four possible outcames are all equiprobable, which is not the case.

Expecially for the first step:If we assumed all the possible split of 49 stalks to be equiprobable, the chance to get a 2 would be

^{11}/

_{47}, meaning that getting a 6 as final outcome would have a probability of 1,28% which is much lower than

^{1}/

_{16}(6.25%).

However, how "random" would you consider a split where one group would contain just one stalk? Not much, I guess.

In fact, the closer the 49 stalks are split

*in the middle*, the closer the chance of getting 2 approximates

^{1}/

_{4 }(and hence the probability of getting 6 get closer to

^{1}/

_{16}).

This is true for the second and third subdivision as well, but the effect is not very relevant and the probability to get 2 or 3 are really

^{1}/

_{2 .}

This leads to the interesting, at least for me, conclusion that we cannot tell the exact probability distribution of the yarrow stalks method:

With the yarrow stalks method, the probability of getting6,7,8, and9is, respectively,^{1}/_{16},^{5}/_{16},^{7}/_{16}and^{3}/_{16}only in theidealcase of splitting the stalks very close to the middle of the heap. The actual probabilities depend on the way the splits are done and varies, even if only slightly, at each cast.

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